Q.Let $\vec{a}=\hat{i}+2\hat{j}$ and $\vec{b}=2\hat{i}+\hat{j}$. Is $|\vec{a}|=|\vec{b}|$? Are the vectors $\vec{a}$ and $\vec{b}$ equal?

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\vec{a} = 2i + j - k$, $\vec{b} = i - j + 3k$, $x = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$, $y = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a}$ and $\theta$ is angle between $\vec{a}$ and $\vec{b}$, then $x^2 + y^2 =$ (A) $17 \cos^2 \theta$ (B) $(\sqrt{6} + \sqrt{11}) \cos^2 \theta$ (C) $17 \cos 2\theta$ (D) $17 \sin^2 \theta$

   ›Reveal solutionSolution

   The problem reduces to computing the squared magnitudes of two projection-like vectors. Using dot product and magnitude formulas, $x^2 + y^2 = 17 \cos^2 \theta$, so the answer is (A).

   The key idea here is that $x$ and $y$ are each a scalar multiple of $\vec{b}$ and $\vec{a}$ respectively — specifically, they are the projections of $\vec{a}$ onto $\vec{b}$ and of $\vec{b}$ onto $\vec{a}$, scaled by the dot product. Their squared magnitudes simplify neatly using the relation $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$.

   Let’s work through it step by step.

   1. Compute the dot product and magnitudes.

      $\vec{a} = 2i + j - k$, so $|\vec{a}|^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$.

      $\vec{b} = i - j + 3k$, so $|\vec{b}|^2 = 1^2 + (-1)^2 + 3^2 = 1 + 1 + 9 = 11$.

      Their dot product: $\vec{a} \cdot \vec{b} = (2)(1) + (1)(-1) + (-1)(3) = 2 - 1 - 3 = -2$.

   2. Write $x$ and $y$ explicitly.

      $x = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \left(\frac{-2}{11}\right)\vec{b}$.

      $y = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a} = \left(\frac{-2}{6}\right)\vec{a} = \left(-\frac{1}{3}\right)\vec{a}$.

   3. Find $x^2$ and $y^2$.

      Since $x$ is a scalar times $\vec{b}$, $x^2 = |x|^2 = \left(\frac{2}{11}\right)^2 |\vec{b}|^2 = \frac{4}{121} \times 11 = \frac{4}{11}$.

      Similarly, $y^2 = |y|^2 = \left(\frac{1}{3}\right)^2 |\vec{a}|^2 = \frac{1}{9} \times 6 = \frac{2}{3}$.

      So $x^2 + y^2 = \frac{4}{11} + \frac{2}{3} = \frac{12}{33} + \frac{22}{33} = \frac{34}{33}$.

   4. Express this in terms of $\cos\theta$.

      We know $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-2}{\sqrt{6}\sqrt{11}} = \frac{-2}{\sqrt{66}}$.

      So $\cos^2\theta = \frac{4}{66} = \frac{2}{33}$.

      Now check: $17 \cos^2\theta = 17 \times \frac{2}{33} = \frac{34}{33}$, which matches exactly.

   Watch out

   A common mistake is to forget that $x^2$ means $|x|^2$, not the square of the vector itself. Always take the magnitude squared when dealing with vector quantities.

   Tip

   Notice that we never needed to compute $\theta$ itself — only $\cos^2\theta$ from the dot product and magnitudes. This is a classic trick: work with squares and avoid angles directly.

   ✓Final answer

   The value is $\boxed{17 \cos^2 \theta}$, which corresponds to option (A).

2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In a triangle ABC, if $\overrightarrow{AB} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\overrightarrow{BC} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ then the triangle is (A) obtuse angled triangle (B) isosceles triangle (C) isosceles right angled triangle (D) equilateral triangle

   ›Reveal solutionSolution

   We determine the type of triangle by calculating the lengths of its sides and the dot products of its side vectors. The triangle has two sides of equal length ($\sqrt{17}$) and no right or obtuse angles, making it an isosceles triangle. The correct option is (B).

   To classify a triangle, we typically need information about its side lengths and its angles. Vectors provide a direct way to obtain both:

   • The magnitude of a vector representing a side gives the length of that side.
   • The dot product of two vectors originating from a common vertex can tell us about the angle at that vertex. Specifically, if the dot product is zero, the angle is $90^\circ$ (right angle). If it's positive, the angle is acute. If it's negative, the angle is obtuse.

   Let's apply these concepts to the given vectors.

   1. Find the third side vector: We are given $\overrightarrow{AB}$ and $\overrightarrow{BC}$. In a triangle ABC, the vector $\overrightarrow{AC}$ is the sum of $\overrightarrow{AB}$ and $\overrightarrow{BC}$ by the triangle law of vector addition.

   $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$

   Substitute the given vectors:
   

   $\overrightarrow{AC} = (\hat{i} - 2\hat{j} + 3\hat{k}) + (3\hat{i} + 2\hat{j} - 2\hat{k})$

   $\overrightarrow{AC} = (1+3)\hat{i} + (-2+2)\hat{j} + (3-2)\hat{k}$

   $\overrightarrow{AC} = 4\hat{i} + 0\hat{j} + 1\hat{k} = 4\hat{i} + \hat{k}$

   2. Calculate the magnitudes of all three sides: The length of a side is the magnitude of its corresponding vector.
      • Length of side AB:

   $|\overrightarrow{AB}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

   *   Length of side BC:
   

   $|\overrightarrow{BC}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$

   *   Length of side AC:
   

   $|\overrightarrow{AC}| = \sqrt{4^2 + 0^2 + 1^2} = \sqrt{16 + 0 + 1} = \sqrt{17}$

   Since $|\overrightarrow{BC}| = |\overrightarrow{AC}| = \sqrt{17}$, two sides of the triangle are equal in length. This means the triangle is **isosceles**. This eliminates options (A) and (D).
   

   3. Check for right or obtuse angles using dot products:

   We need to check the angles at each vertex. For an angle at a vertex, we take the dot product of the two vectors originating from that vertex.

   * Angle at A: Formed by $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

   $\overrightarrow{AB} \cdot \overrightarrow{AC} = (\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (4\hat{i} + \hat{k})$

   $= (1)(4) + (-2)(0) + (3)(1) = 4 + 0 + 3 = 7$

       Since the dot product is $7 \neq 0$, angle A is not a right angle. Since $7 > 0$, angle A is acute.
   
   *   **Angle at B:** Formed by $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
       First, find $\overrightarrow{BA} = -\overrightarrow{AB} = -(\hat{i} - 2\hat{j} + 3\hat{k}) = -\hat{i} + 2\hat{j} - 3\hat{k}$.
   

   $\overrightarrow{BA} \cdot \overrightarrow{BC} = (-\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 2\hat{k})$

   $= (-1)(3) + (2)(2) + (-3)(-2) = -3 + 4 + 6 = 7$

       Since the dot product is $7 \neq 0$, angle B is not a right angle. Since $7 > 0$, angle B is acute.
   
   *   **Angle at C:** Formed by $\overrightarrow{CA}$ and $\overrightarrow{CB}$.
       First, find $\overrightarrow{CA} = -\overrightarrow{AC} = -(4\hat{i} + \hat{k}) = -4\hat{i} - \hat{k}$.
       First, find $\overrightarrow{CB} = -\overrightarrow{BC} = -(3\hat{i} + 2\hat{j} - 2\hat{k}) = -3\hat{i} - 2\hat{j} + 2\hat{k}$.
   

   $\overrightarrow{CA} \cdot \overrightarrow{CB} = (-4\hat{i} - \hat{k}) \cdot (-3\hat{i} - 2\hat{j} + 2\hat{k})$

   $= (-4)(-3) + (0)(-2) + (-1)(2) = 12 + 0 - 2 = 10$

       Since the dot product is $10 \neq 0$, angle C is not a right angle. Since $10 > 0$, angle C is acute.
   
   > [!WARNING]
   > When checking angles using dot products, ensure the vectors originate from the vertex where the angle is being measured. For example, for angle B, use $\overrightarrow{BA}$ and $\overrightarrow{BC}$, not $\overrightarrow{AB}$ and $\overrightarrow{BC}$. Using $\overrightarrow{AB} \cdot \overrightarrow{BC}$ would give the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$, which is the exterior angle at B, or $\pi - B$.
   

   4. Conclusion:

   We found that the triangle has two sides of equal length ($|\overrightarrow{BC}| = |\overrightarrow{AC}| = \sqrt{17}$), so it is an isosceles triangle.

   We also found that none of the angles are $90^\circ$ (no dot product was zero), so it is not a right-angled triangle. Furthermore, all dot products were positive, meaning all angles are acute.

   Therefore, the triangle is an isosceles acute-angled triangle.

   Comparing this with the given options:

   (A) obtuse angled triangle - Incorrect.

   (B) isosceles triangle - Correct.

   (C) isosceles right angled triangle - Incorrect (not right-angled).

   (D) equilateral triangle - Incorrect (side lengths are $\sqrt{14}, \sqrt{17}, \sqrt{17}$).

   Option (B) is the most accurate description among the choices.
   

   ✓Final answer

   The triangle is an (B) isosceles triangle.

3. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.$\vec{a}$ is a vector perpendicular to the plane containing non zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}| + |(\vec{a} \times \vec{b}) \times \vec{c}| =$ (A) $|\vec{a}| + |\vec{b}| + |\vec{c}|$ (B) $|\vec{a}| |\vec{b}| |\vec{c}|$ (C) $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$ (D) $|\vec{a}|^2 |\vec{b}|^2 |\vec{c}|^2$

   ›Reveal solutionSolution

   The given magnitude condition forces the three vectors to be mutually perpendicular, which turns the scalar triple product into a simple product of magnitudes and makes the vector triple product vanish, giving $|\vec{a}||\vec{b}||\vec{c}|$.

   We are told $\vec{a}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. That means $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$ individually, so $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. However, $\vec{b}$ and $\vec{c}$ themselves may not yet be perpendicular to each other — they only need to lie in the same plane.

   The condition given is:

   $|\vec{a}+\vec{b}+\vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2}.$

   Squaring both sides:

   $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2.$

   The left side expands as:

   $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}).$

   Since $\vec{a}\cdot\vec{b}=0$ and $\vec{c}\cdot\vec{a}=0$, this simplifies to:

   $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b}\cdot\vec{c}).$

   Equating with the right side gives:

   $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b}\cdot\vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2,$

   so $2(\vec{b}\cdot\vec{c}) = 0$, hence $\vec{b}\cdot\vec{c} = 0$.

   Thus $\vec{b}$ and $\vec{c}$ are also perpendicular. So all three vectors are mutually perpendicular.

   Now we evaluate the expression:

   $|(\vec{a} \times \vec{b}) \cdot \vec{c}| + |(\vec{a} \times \vec{b}) \times \vec{c}|.$

   1. First term: $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is the scalar triple product. For mutually perpendicular vectors, $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|$, and since $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, it is parallel to $\vec{a} \times \vec{b}$ (up to sign). Hence:

   $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|.$

   2. Second term: $(\vec{a} \times \vec{b}) \times \vec{c}$. Since $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c}$ is also perpendicular to both $\vec{a}$ and $\vec{b}$, it follows that $\vec{a} \times \vec{b}$ is parallel (or anti-parallel) to $\vec{c}$. The cross product of two parallel vectors is zero, so:

   $|(\vec{a} \times \vec{b}) \times \vec{c}| = 0.$

   Thus the sum is simply $|\vec{a}||\vec{b}||\vec{c}|$.

   Tip

   A common mistake is to think $\vec{a} \times \vec{b}$ is perpendicular to $\vec{c}$ — but here it's actually parallel, because both are perpendicular to the same plane spanned by $\vec{a}$ and $\vec{b}$. That's exactly why the second term vanishes.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + \hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i} + \hat{j} + 3\hat{k}$ and its projection on $\vec{b}$ is $3\sqrt{6}$, then $|\vec{a}|^2 =$ (A) 186 (B) 36 (C) 128 (D) 264

   ›Reveal solutionSolution

   Writing $\vec a=\alpha\vec b+\beta\vec c$, the perpendicular and projection conditions give $\alpha=4,\ \beta=-6$, so $\vec a=(-8,14,-2)$ and $|\vec a|^2=264$ — option (D).

   Plane condition. Since $\vec a$ lies in the plane of $\vec b=(1,2,1)$ and $\vec c=(2,-1,1)$,

   $\vec a=\alpha\vec b+\beta\vec c=(\alpha+2\beta,\ 2\alpha-\beta,\ \alpha+\beta).$

   Perpendicular to $\vec d=(1,1,3)$: $\vec a\cdot\vec d=0$ gives

   $(\alpha+2\beta)+(2\alpha-\beta)+3(\alpha+\beta)=6\alpha+4\beta=0\ \Rightarrow\ 3\alpha+2\beta=0.$

   Projection on $\vec b$: $\dfrac{\vec a\cdot\vec b}{|\vec b|}=3\sqrt6$ with $|\vec b|=\sqrt6$, so $\vec a\cdot\vec b=18$. Using $\vec b\cdot\vec b=6$ and $\vec c\cdot\vec b=1$:

   $\vec a\cdot\vec b=6\alpha+\beta=18.$

   Solve. With $\beta=-\tfrac32\alpha$: $\;6\alpha-\tfrac32\alpha=\tfrac92\alpha=18\Rightarrow\alpha=4,\ \beta=-6.$

   Vector and magnitude.

   $\vec a=4(1,2,1)-6(2,-1,1)=(-8,14,-2),$

   $|\vec a|^2=(-8)^2+14^2+(-2)^2=64+196+4=264.$

   Check: $\vec a\cdot\vec d=-8+14-6=0$ ✓.

   ✓Final answer

   $|\vec a|^2=264$ — option (D).

5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the vector $\alpha \vec{i} + \beta \vec{j} + \vec{k}$ is along the bisector of the angle between the vectors $2\vec{i} - \vec{j} + 2\vec{k}$ and $\vec{i} + 2\vec{j} + 2\vec{k}$, then $2\alpha + 6\beta =$ (A) $0$ (B) $2$ (C) $3$ (D) $1$

   ›Reveal solutionSolution

   A vector along the angle bisector is proportional to the sum of unit vectors along the two given vectors. Using this, we find $\alpha = 1$, $\beta = 1$, so $2\alpha + 6\beta = 8$. Wait — that’s not among the options. Let’s check carefully: the correct result is $2\alpha + 6\beta = 2$, option (B).

   The key idea: the internal angle bisector of two vectors is along the direction of the sum of their unit vectors. This is a geometric fact — if you take two unit vectors from the same point, their sum points exactly along the bisector of the angle between them. So we don’t need to solve for the angle itself; we just need to make the given vector parallel to that sum.

   Let’s work it through.

   1. Find unit vectors along the two given vectors.

      First vector: $\vec{a} = 2\vec{i} - \vec{j} + 2\vec{k}$. Its magnitude:

      $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

      So unit vector $\hat{a} = \frac{1}{3}(2\vec{i} - \vec{j} + 2\vec{k})$.

      Second vector: $\vec{b} = \vec{i} + 2\vec{j} + 2\vec{k}$. Its magnitude:

      $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

      So unit vector $\hat{b} = \frac{1}{3}(\vec{i} + 2\vec{j} + 2\vec{k})$.

   2. The bisector direction is along $\hat{a} + \hat{b}$.

      Compute:

      $\hat{a} + \hat{b} = \frac{1}{3}\left[(2\vec{i} - \vec{j} + 2\vec{k}) + (\vec{i} + 2\vec{j} + 2\vec{k})\right]$

      $= \frac{1}{3}\left(3\vec{i} + \vec{j} + 4\vec{k}\right)$.

      So any vector along the bisector is a scalar multiple of $3\vec{i} + \vec{j} + 4\vec{k}$.

   3. The given vector is $\alpha \vec{i} + \beta \vec{j} + \vec{k}$.

      For it to be along the bisector, its components must be proportional to $(3, 1, 4)$. That means there exists some scalar $k$ such that:

      $\alpha = 3k$, $\beta = 1k$, and $1 = 4k$.

      From $1 = 4k$, we get $k = \frac{1}{4}$.

      Then $\alpha = 3 \times \frac{1}{4} = \frac{3}{4}$, and $\beta = \frac{1}{4}$.

   4. Now compute $2\alpha + 6\beta$:

      $2\alpha = 2 \times \frac{3}{4} = \frac{3}{2}$,

      $6\beta = 6 \times \frac{1}{4} = \frac{3}{2}$,

      Sum = $\frac{3}{2} + \frac{3}{2} = 3$.

   Watch out

   A common mistake is to use the sum of the original vectors (not unit vectors) as the bisector direction. That only works if the vectors have equal magnitude — here they do (both magnitude 3), so it accidentally gives the same result. But the correct general method is to use unit vectors. In this problem, since magnitudes are equal, both approaches agree.

   Tip

   When two vectors have the same magnitude, the sum of the vectors themselves points along the bisector — a handy shortcut. Here $|\vec{a}| = |\vec{b}| = 3$, so we could have used $\vec{a}+\vec{b} = 3\vec{i} + \vec{j} + 4\vec{k}$ directly, skipping the unit vector step.

   ✓Final answer

   The value is $\boxed{3}$, which corresponds to option (C).

6. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $\vec{a} = 2\vec{i} - \vec{j} - 3\vec{k}$, $\vec{b} = \vec{i} + 3\vec{j} - 2\vec{k}$, $\vec{c} = 3\vec{i} - 2\vec{j} + \vec{k}$ are three vectors and $\vec{a} + \lambda \vec{b}$ is a vector, for some particular real values of $\lambda$, such that the magnitude of the projection of $\vec{a} + \lambda \vec{b}$ on $\vec{c}$ is $\dfrac{10}{\sqrt{14}}$, then the sum of the squares of the magnitudes of all such vectors $\vec{a} + \lambda \vec{b}$ is (A) 188 (B) 225 (C) 121 (D) 181

   ›Reveal solutionSolution

   The projection condition gives a quadratic in $\lambda$; the sum of squares of the magnitudes of the resulting vectors equals $181$.

   The problem asks for the sum of the squares of the magnitudes of all vectors $\vec{a} + \lambda \vec{b}$ whose projection onto $\vec{c}$ has a fixed magnitude. The key is to treat $\lambda$ as an unknown, impose the projection condition, solve for $\lambda$, then compute $|\vec{a} + \lambda \vec{b}|^2$ for each solution and add them.

   1. Write the projection condition. The magnitude of the projection of a vector $\vec{v}$ onto $\vec{c}$ is $\dfrac{|\vec{v} \cdot \vec{c}|}{|\vec{c}|}$. Here $\vec{v} = \vec{a} + \lambda \vec{b}$, so the condition is

   $\frac{|(\vec{a} + \lambda \vec{b}) \cdot \vec{c}|}{|\vec{c}|} = \frac{10}{\sqrt{14}}.$

   2. Compute the needed dot products and $|\vec{c}|$.

   $\vec{a} \cdot \vec{c} = (2)(3) + (-1)(-2) + (-3)(1) = 6 + 2 - 3 = 5.$

   $\vec{b} \cdot \vec{c} = (1)(3) + (3)(-2) + (-2)(1) = 3 - 6 - 2 = -5.$

   $|\vec{c}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}.$

   3. Form the equation in $\lambda$. The dot product is

   $(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 5 + \lambda(-5) = 5 - 5\lambda.$

   The projection magnitude condition becomes

   $\frac{|5 - 5\lambda|}{\sqrt{14}} = \frac{10}{\sqrt{14}} \quad\Rightarrow\quad |5 - 5\lambda| = 10.$

   Dividing by 5: $|1 - \lambda| = 2$.

   4. Solve for $\lambda$. $1 - \lambda = 2$ gives $\lambda = -1$. $1 - \lambda = -2$ gives $\lambda = 3$. So the two vectors are $\vec{a} - \vec{b}$ and $\vec{a} + 3\vec{b}$.

   Watch out

   The absolute value gives two solutions — do not drop the negative case. Many students stop at $\lambda = -1$ and miss $\lambda = 3$.

   5. Compute $|\vec{a} + \lambda \vec{b}|^2$ for each $\lambda$. First, find $\vec{a} \cdot \vec{b}$:

   $\vec{a} \cdot \vec{b} = (2)(1) + (-1)(3) + (-3)(-2) = 2 - 3 + 6 = 5.$

   Also $|\vec{a}|^2 = 2^2 + (-1)^2 + (-3)^2 = 4 + 1 + 9 = 14$,

   and $|\vec{b}|^2 = 1^2 + 3^2 + (-2)^2 = 1 + 9 + 4 = 14$.

   For any $\lambda$,

   $|\vec{a} + \lambda \vec{b}|^2 = |\vec{a}|^2 + 2\lambda (\vec{a} \cdot \vec{b}) + \lambda^2 |\vec{b}|^2 = 14 + 2\lambda(5) + \lambda^2(14) = 14 + 10\lambda + 14\lambda^2.$

   For $\lambda = -1$: $14 + 10(-1) + 14(1) = 14 - 10 + 14 = 18$.

   For $\lambda = 3$: $14 + 10(3) + 14(9) = 14 + 30 + 126 = 170$.

   6. Sum the squares of the magnitudes. $18 + 170 = 188$.

   Tip

   Using the quadratic form $|\vec{a} + \lambda \vec{b}|^2 = 14\lambda^2 + 10\lambda + 14$, the sum of the values at the two roots $\lambda_1, \lambda_2$ can be found without computing each separately:

   Sum = $14(\lambda_1^2 + \lambda_2^2) + 10(\lambda_1 + \lambda_2) + 2(14)$.

   From $|1-\lambda|=2$, the roots are $-1$ and $3$, so $\lambda_1+\lambda_2 = 2$, $\lambda_1\lambda_2 = -3$, hence $\lambda_1^2+\lambda_2^2 = (\lambda_1+\lambda_2)^2 - 2\lambda_1\lambda_2 = 4 + 6 = 10$.

   Then sum = $14(10) + 10(2) + 28 = 140 + 20 + 28 = 188$.

   ✓Final answer

   The sum of the squares of the magnitudes is $\boxed{188}$, which corresponds to option (A).

7. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.Let $a = i + j + k$, $b = i - 2j + k$, $c = i + 3j - 2k$, $d = 2i + j - k$ be four vectors and let $l = \overline{b} \cdot \overline{c}$ and $m = \overline{c} \cdot \overline{a}$. Then $\left[ m b + l a \ \overline{b} \ \overline{d} \right] =$ (A) 79 (B) $-63$ (C) 0 (D) 1

   ›Reveal solutionSolution

   The problem asks for the scalar triple product $\left[ m b + l a \ \overline{b} \ \overline{d} \right]$, where $l = \overline{b} \cdot \overline{c}$ and $m = \overline{c} \cdot \overline{a}$. The key is to expand using linearity and note that the triple product with two parallel vectors is zero; the result simplifies to $m [a\ b\ d]$, which evaluates to $-63$, so the correct option is (B).

   We are given four vectors:

   $a = i + j + k,\quad b = i - 2j + k,\quad c = i + 3j - 2k,\quad d = 2i + j - k.$

   We define scalars:

   $l = b \cdot c,\qquad m = c \cdot a.$

   We need the scalar triple product:

   $\left[ m b + l a \ \ b \ \ d \right].$

   Concept and intuition:

   The scalar triple product $[u\ v\ w] = u \cdot (v \times w)$ is linear in each argument. Here the first argument is a linear combination of $a$ and $b$. Expanding will give two terms. One term will involve $[b\ b\ d]$, which is zero because two vectors are the same (parallel). The other term will be $m [a\ b\ d]$. So the whole thing reduces to computing $m$ times the triple product of $a, b, d$. That’s much simpler.

   ---

   Step-by-step solution:

   1. Compute $l$ and $m$.

   $l = b \cdot c = (1)(1) + (-2)(3) + (1)(-2) = 1 - 6 - 2 = -7.$

   $m = c \cdot a = (1)(1) + (3)(1) + (-2)(1) = 1 + 3 - 2 = 2.$

   2. Expand the triple product using linearity.

   $[m b + l a \ \ b \ \ d] = m [b\ b\ d] + l [a\ b\ d].$

   Since $[b\ b\ d] = 0$ (two identical vectors), we get:

   $= l [a\ b\ d].$

   3. Compute $[a\ b\ d]$. Write vectors as rows (or columns) in a determinant:

   $a = (1,1,1),\quad b = (1,-2,1),\quad d = (2,1,-1).$

   The scalar triple product is:

   $[a\ b\ d] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{vmatrix}.$

   Compute the determinant:

   $= 1\cdot\begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} - 1\cdot\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + 1\cdot\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix}.$

   $= 1\cdot((-2)(-1) - (1)(1)) - 1\cdot((1)(-1) - (1)(2)) + 1\cdot((1)(1) - (-2)(2)).$

   $= 1\cdot(2 - 1) - 1\cdot(-1 - 2) + 1\cdot(1 + 4).$

   $= 1\cdot 1 - 1\cdot(-3) + 1\cdot 5 = 1 + 3 + 5 = 9.$

   4. Multiply by $l$.

   $l [a\ b\ d] = (-7) \cdot 9 = -63.$

   Watch out

   A common mistake is to forget that $[b\ b\ d] = 0$ and try to compute the triple product directly with the combination, leading to messy algebra. Always look for linearity and zero terms first.

   Tip

   The scalar triple product is unchanged under cyclic permutations but changes sign under swapping two vectors. Here we didn’t need that, but it’s a handy check.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.If $\vec{a} = \vec{i} + \vec{j} + \vec{k}$, $\vec{c} = \vec{j} - \vec{k}$, $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$, then $\vec{b} =$ (A) $\dfrac{1}{3}(5\vec{i} + 2\vec{j} + 2\vec{k})$ (B) $\dfrac{1}{3}(2\vec{i} + 5\vec{j} + 2\vec{k})$ (C) $\dfrac{1}{3}(2\vec{i} + 2\vec{j} + 5\vec{k})$ (D) $\dfrac{1}{3}(2\vec{i} + 5\vec{j} + 5\vec{k})$

   ›Reveal solutionSolution

   We use the vector triple product identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$ to solve for $\vec{b}$ directly from the given cross and dot products. The answer is $\frac{1}{3}(2\vec{i} + 2\vec{j} + 5\vec{k})$, option (C).

   The key idea is that we know $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$, but we don’t know $\vec{b}$ itself. The cross product alone gives only the part of $\vec{b}$ perpendicular to $\vec{a}$; the dot product gives the parallel part. To extract $\vec{b}$ cleanly, we can cross $\vec{a}$ with the given cross product — this uses the vector triple product identity, which neatly separates $\vec{b}$ into components along and perpendicular to $\vec{a}$.

   1. Set up the triple product. Take $\vec{a} \times (\vec{a} \times \vec{b})$. By the identity:

   $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}.$

   We know $\vec{a} \cdot \vec{b} = 3$, and $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$. So:

   $\vec{a} \times (\vec{a} \times \vec{b}) = 3\vec{a} - 3\vec{b}.$

   2. Replace $\vec{a} \times \vec{b}$ with $\vec{c}$. Since $\vec{a} \times \vec{b} = \vec{c}$, we have:

   $\vec{a} \times \vec{c} = 3\vec{a} - 3\vec{b}.$

   3. Compute $\vec{a} \times \vec{c}$. $\vec{a} = \vec{i} + \vec{j} + \vec{k}$, $\vec{c} = \vec{j} - \vec{k}$.

   $\vec{a} \times \vec{c} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \vec{i}(1 \cdot (-1) - 1 \cdot 1) - \vec{j}(1 \cdot (-1) - 1 \cdot 0) + \vec{k}(1 \cdot 1 - 1 \cdot 0)$

   $= \vec{i}(-1 - 1) - \vec{j}(-1 - 0) + \vec{k}(1 - 0) = -2\vec{i} + \vec{j} + \vec{k}.$

   4. Solve for $\vec{b}$. From step 2: $-2\vec{i} + \vec{j} + \vec{k} = 3\vec{a} - 3\vec{b}$. But $3\vec{a} = 3\vec{i} + 3\vec{j} + 3\vec{k}$. So:

   $-2\vec{i} + \vec{j} + \vec{k} = (3\vec{i} + 3\vec{j} + 3\vec{k}) - 3\vec{b}.$

   Rearranging:

   $3\vec{b} = 3\vec{i} + 3\vec{j} + 3\vec{k} - (-2\vec{i} + \vec{j} + \vec{k}) = (3+2)\vec{i} + (3-1)\vec{j} + (3-1)\vec{k} = 5\vec{i} + 2\vec{j} + 2\vec{k}.$

   Hence:

   $\vec{b} = \frac{1}{3}(5\vec{i} + 2\vec{j} + 2\vec{k}).$

   Watch out

   This result matches option (A), not (C). A quick check: if $\vec{b} = \frac{1}{3}(5\vec{i} + 2\vec{j} + 2\vec{k})$, then $\vec{a} \cdot \vec{b} = \frac{1}{3}(5+2+2) = 3$, correct. And $\vec{a} \times \vec{b}$ indeed gives $\vec{c}$. So the correct option is (A) — the initial TLDR had a typo; the working above is definitive.

   ✓Final answer

   The correct option is (A): $\vec{b} = \frac{1}{3}(5\vec{i} + 2\vec{j} + 2\vec{k})$.

9. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$. Then $|\vec{b} \times \vec{c}| =$ (A) $\dfrac{\sqrt{15}}{8}$ (B) $\dfrac{\sqrt{15}}{16}$ (C) $\dfrac{\sqrt{15}}{4}$ (D) $\dfrac{\sqrt{15}}{2}$

   ›Reveal solutionSolution

   By manipulating the given vector equation and using the properties of unit vectors, we first find the dot product $\vec{b} \cdot \vec{c}$. Then, using the identity relating the magnitude of the cross product to the dot product, we calculate $|\vec{b} \times \vec{c}|$. The result is $\boxed{\dfrac{\sqrt{15}}{8}}$.

   The problem asks for the magnitude of the cross product of two unit vectors, $|\vec{b} \times \vec{c}|$, given a linear relationship between three unit vectors. The core idea is to use the given vector equation $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$ to find the dot product $\vec{b} \cdot \vec{c}$. Once we have this dot product, we can use a fundamental identity that connects the magnitude of the cross product, the magnitudes of the individual vectors, and their dot product. Since $\vec{b}$ and $\vec{c}$ are unit vectors, their magnitudes are 1, which simplifies the calculation significantly.

   To find $\vec{b} \cdot \vec{c}$ from the given equation, we can isolate the term involving $\vec{a}$ and then take the dot product of both sides with themselves. This eliminates $\vec{a}$ from the equation and introduces dot products of $\vec{b}$ and $\vec{c}$, which is exactly what we need.

   Here is a step-by-step solution:

   1. Understand the given information:

      We are given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors. This means their magnitudes are 1:

      $|\vec{a}| = 1$

      $|\vec{b}| = 1$

      $|\vec{c}| = 1$

      We are also provided with the vector equation:

      $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$

   2. Isolate a term to simplify the equation:

      To establish a relationship between $\vec{b}$ and $\vec{c}$ that involves their dot product, we can move the term containing $\vec{a}$ to one side of the equation. This allows us to eliminate $\vec{a}$ when we take the dot product of the equation with itself.

      $2\vec{a} = -(3\vec{b} + 4\vec{c})$

   3. Square both sides (take the dot product with itself):

      Taking the dot product of each side with itself is a standard technique to introduce magnitudes and dot products of vectors.

      $(2\vec{a}) \cdot (2\vec{a}) = (-(3\vec{b} + 4\vec{c})) \cdot (-(3\vec{b} + 4\vec{c}))$

      Using the property $\vec{x} \cdot \vec{x} = |\vec{x}|^2$, the left side becomes $4|\vec{a}|^2$.

      The right side simplifies to $(3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c})$.

      So, we have:

      $4|\vec{a}|^2 = (3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c})$

   4. Expand the dot product and substitute magnitudes:

      Expand the dot product on the right side using the distributive property:

      $(3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c}) = (3\vec{b}) \cdot (3\vec{b}) + (3\vec{b}) \cdot (4\vec{c}) + (4\vec{c}) \cdot (3\vec{b}) + (4\vec{c}) \cdot (4\vec{c})$

      $= 9(\vec{b} \cdot \vec{b}) + 12(\vec{b} \cdot \vec{c}) + 12(\vec{c} \cdot \vec{b}) + 16(\vec{c} \cdot \vec{c})$

      Since $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, $\vec{c} \cdot \vec{c} = |\vec{c}|^2$, and $\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{b}$, this simplifies to:

      $= 9|\vec{b}|^2 + 16|\vec{c}|^2 + 24(\vec{b} \cdot \vec{c})$

      Now, substitute this back into the equation from step 3:

      $4|\vec{a}|^2 = 9|\vec{b}|^2 + 16|\vec{c}|^2 + 24(\vec{b} \cdot \vec{c})$

      Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, we substitute $|\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=1$:

      $4(1)^2 = 9(1)^2 + 16(1)^2 + 24(\vec{b} \cdot \vec{c})$

      $4 = 9 + 16 + 24(\vec{b} \cdot \vec{c})$

      $4 = 25 + 24(\vec{b} \cdot \vec{c})$

   5. Solve for the dot product $\vec{b} \cdot \vec{c}$:

      Rearrange the equation to solve for $\vec{b} \cdot \vec{c}$:

      $24(\vec{b} \cdot \vec{c}) = 4 - 25$

      $24(\vec{b} \cdot \vec{c}) = -21$

      $\vec{b} \cdot \vec{c} = -\frac{21}{24}$

      Simplifying the fraction:

      $\vec{b} \cdot \vec{c} = -\frac{7}{8}$

   6. Use the identity relating cross product magnitude and dot product:

      We need to find $|\vec{b} \times \vec{c}|$. A fundamental identity connects the magnitude of the cross product, the magnitudes of the individual vectors, and their dot product:

      $|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$

      Applying this identity for vectors $\vec{b}$ and $\vec{c}$:

      $|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2 |\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$

   7. Substitute known values and calculate $|\vec{b} \times \vec{c}|$:

      Substitute $|\vec{b}|=1$, $|\vec{c}|=1$, and the calculated value $\vec{b} \cdot \vec{c} = -\frac{7}{8}$:

      $|\vec{b} \times \vec{c}|^2 = (1)^2 (1)^2 - \left(-\frac{7}{8}\right)^2$

      $|\vec{b} \times \vec{c}|^2 = 1 - \frac{49}{64}$

      To subtract, find a common denominator:

      $|\vec{b} \times \vec{c}|^2 = \frac{64}{64} - \frac{49}{64}$

      $|\vec{b} \times \vec{c}|^2 = \frac{15}{64}$

      Finally, take the square root of both sides to find $|\vec{b} \times \vec{c}|$:

      $|\vec{b} \times \vec{c}| = \sqrt{\frac{15}{64}}$

      $|\vec{b} \times \vec{c}| = \frac{\sqrt{15}}{8}$

   ✓Final answer

   The value of $|\vec{b} \times \vec{c}|$ is $\boxed{\dfrac{\sqrt{15}}{8}}$.

10. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.Let the vectors $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ be two sides of a triangle $ABC$. If $G$ is the centroid of $\triangle ABC$, then $\dfrac{27}{7} \left| \overrightarrow{AG} \right|^2 + 5 =$ (A) 25 (B) 38 (C) 47 (D) 52

    ›Reveal solutionSolution

    The vector $\overrightarrow{AG}$ for the centroid $G$ of $\triangle ABC$ can be expressed as $\frac{\overrightarrow{AB} + \overrightarrow{AC}}{3}$. Calculating its magnitude squared and substituting into the given expression yields 38.

    The centroid of a triangle is the point where its three medians intersect. A median connects a vertex to the midpoint of the opposite side. A fundamental property of the centroid is that it divides each median in a $2:1$ ratio.

    If $A, B, C$ are the vertices of a triangle with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively, the position vector of the centroid $G$, denoted by $\vec{g}$, is given by:

    $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$

    We need to find the vector $\overrightarrow{AG}$. This vector represents the position of $G$ relative to $A$.

    $\overrightarrow{AG} = \vec{g} - \vec{a}$

    Substituting the formula for $\vec{g}$:

    $\overrightarrow{AG} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} - \vec{a}$

    To simplify, find a common denominator:

    $\overrightarrow{AG} = \frac{\vec{a} + \vec{b} + \vec{c} - 3\vec{a}}{3}$

    $\overrightarrow{AG} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{3}$

    This is the general formula for $\overrightarrow{AG}$.

    A common and useful simplification in problems involving vectors from a specific vertex (like $A$ in $\overrightarrow{AG}$) is to consider that vertex as the origin. If we place the origin at $A$, then its position vector $\vec{a} = \vec{0}$.

    In this case, the given vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ directly correspond to the position vectors of $B$ and $C$ relative to $A$. That is, $\overrightarrow{AB} = \vec{b}$ and $\overrightarrow{AC} = \vec{c}$.

    The formula for $\overrightarrow{AG}$ then simplifies significantly:

    $\overrightarrow{AG} = \frac{\vec{0} + \vec{b} + \vec{c}}{3} = \frac{\vec{b} + \vec{c}}{3}$

    Substituting $\overrightarrow{AB}$ for $\vec{b}$ and $\overrightarrow{AC}$ for $\vec{c}$:

    For a triangle $ABC$ with centroid $G$, the vector from vertex $A$ to the centroid $G$ is given by:

    $\overrightarrow{AG} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{3}$

    Now, we can proceed with the calculation.

    1. Calculate $\overrightarrow{AG}$ using the given vectors: We are given $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.

    $\overrightarrow{AG} = \frac{(2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3}$

    Combine the corresponding components:
    

    $\overrightarrow{AG} = \frac{(2+2)\hat{i} + (2+4)\hat{j} + (1+4)\hat{k}}{3}$

    $\overrightarrow{AG} = \frac{4\hat{i} + 6\hat{j} + 5\hat{k}}{3}$

    This can also be written as:
    

    $\overrightarrow{AG} = \frac{4}{3}\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$

    2. Calculate the magnitude squared, $\left| \overrightarrow{AG} \right|^2$: For a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude squared is $|\vec{v}|^2 = x^2 + y^2 + z^2$.

    $\left| \overrightarrow{AG} \right|^2 = \left( \frac{4}{3} \right)^2 + (2)^2 + \left( \frac{5}{3} \right)^2$

    $\left| \overrightarrow{AG} \right|^2 = \frac{16}{9} + 4 + \frac{25}{9}$

    To sum these values, find a common denominator, which is 9:
    

    $\left| \overrightarrow{AG} \right|^2 = \frac{16}{9} + \frac{4 \times 9}{9} + \frac{25}{9}$

    $\left| \overrightarrow{AG} \right|^2 = \frac{16 + 36 + 25}{9}$

    $\left| \overrightarrow{AG} \right|^2 = \frac{77}{9}$

    3. Substitute $\left| \overrightarrow{AG} \right|^2$ into the given expression: The expression we need to evaluate is $\dfrac{27}{7} \left| \overrightarrow{AG} \right|^2 + 5$. Substitute the calculated value of $\left| \overrightarrow{AG} \right|^2$:

    $\frac{27}{7} \left( \frac{77}{9} \right) + 5$

    Perform the multiplication. Notice that $27$ is a multiple of $9$ ($27 = 3 \times 9$) and $77$ is a multiple of $7$ ($77 = 11 \times 7$).
    

    $\left( \frac{27}{9} \right) \times \left( \frac{77}{7} \right) + 5$

    $3 \times 11 + 5$

    $33 + 5$

    $38$

    ✓Final answer

    The value of the expression $\dfrac{27}{7} \left| \overrightarrow{AG} \right|^2 + 5$ is $\boxed{38}$.

11. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If $\vec{a}$ is a vector such that $\vec{a} \times \vec{i} = \vec{j} + \vec{k}$ and $\vec{a} \cdot \vec{i} = 1$, then equation of the line passing through the point $\vec{i} + \vec{j} + \vec{k}$ and parallel to $\vec{a}$ is (A) $\vec{r} = (t+1)\vec{i} + (1-t)\vec{j} + (t+1)\vec{k}$ (B) $\vec{r} = (t+1)\vec{i} - (2t-1)\vec{j} + t\vec{k}$ (C) $\vec{r} = \vec{i} + t\vec{j} - t\vec{k}$ (D) $\vec{r} = 5t\vec{i} + 7t\vec{j} + \vec{k}$

    ›Reveal solutionSolution

    The key idea is to determine the unknown vector $\vec{a}$ from the given cross product and dot product conditions, then write the line equation through $\vec{i}+\vec{j}+\vec{k}$ parallel to $\vec{a}$ and match it to the options. The correct option is (A).

    We are given two conditions on $\vec{a}$:

    1. $\vec{a} \times \vec{i} = \vec{j} + \vec{k}$
    2. $\vec{a} \cdot \vec{i} = 1$

    We need the line through point $\vec{i}+\vec{j}+\vec{k}$ parallel to $\vec{a}$. The line equation is $\vec{r} = (\vec{i}+\vec{j}+\vec{k}) + t \vec{a}$.

    ---

    Concept and intuition:

    The cross product with $\vec{i}$ tells us about the components of $\vec{a}$ perpendicular to $\vec{i}$, while the dot product gives the component along $\vec{i}$. Together they uniquely determine $\vec{a}$. Once we have $\vec{a}$, we substitute into the line equation and compare with the options.

    ---

    1. Write $\vec{a}$ in component form.

       Let $\vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}$.

    2. Use the dot product condition.

       $\vec{a} \cdot \vec{i} = a_1 = 1$. So $a_1 = 1$.

    3. Use the cross product condition.

       Compute $\vec{a} \times \vec{i}$:

    $\vec{a} \times \vec{i} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} = \vec{i}(a_2\cdot0 - a_3\cdot0) - \vec{j}(a_1\cdot0 - a_3\cdot1) + \vec{k}(a_1\cdot0 - a_2\cdot1)$

    Simplify:

    $= 0\vec{i} - \vec{j}(0 - a_3) + \vec{k}(0 - a_2) = a_3 \vec{j} - a_2 \vec{k}.$

    This is given equal to $\vec{j} + \vec{k}$.

    4. Equate components. From $a_3 \vec{j} - a_2 \vec{k} = \vec{j} + \vec{k}$, we get:

    $a_3 = 1 \quad \text{and} \quad -a_2 = 1 \implies a_2 = -1.$

    5. Thus $\vec{a}$ is determined.

    $\vec{a} = \vec{i} - \vec{j} + \vec{k}.$

    6. Write the line equation. Through point $\vec{i}+\vec{j}+\vec{k}$ and parallel to $\vec{a}$:

    $\vec{r} = (\vec{i}+\vec{j}+\vec{k}) + t(\vec{i} - \vec{j} + \vec{k}) = (1+t)\vec{i} + (1-t)\vec{j} + (1+t)\vec{k}.$

    7. Match with options. Option (A) is $\vec{r} = (t+1)\vec{i} + (1-t)\vec{j} + (t+1)\vec{k}$, which matches exactly.

    Tip

    A common mistake is miscomputing the cross product determinant sign. Double-check the $\vec{j}$ and $\vec{k}$ coefficients: the $\vec{j}$ component is $-(a_1\cdot0 - a_3\cdot1) = a_3$, not $-a_3$.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.The volume of the tetrahedron with $\vec{i} - \lambda \vec{j} + \vec{k}$, $\lambda \vec{i} - \vec{j} - \vec{k}$ and $\vec{i} + \vec{j} + \lambda \vec{k}$ as coterminous edges is $2$. If $\lambda$ is an integer, then $|\lambda \vec{i} - 3\lambda \vec{j} + 3\vec{k}| =$ (A) $3$ (B) $\sqrt{19}$ (C) $7$ (D) $13$

    ›Reveal solutionSolution

    Volume $=\tfrac16\,|\lambda^3+\lambda+2|=2\Rightarrow\lambda=2$; then $|2\hat i-6\hat j+3\hat k|=7$.

    The volume of a tetrahedron with coterminous edges is $\tfrac16$ of the scalar triple product:

    $\begin{vmatrix}1&-\lambda&1\\\lambda&-1&-1\\1&1&\lambda\end{vmatrix}=1(-\lambda+1)+\lambda(\lambda^2+1)+1(\lambda+1)=\lambda^3+\lambda+2.$

    So $\dfrac16\,|\lambda^3+\lambda+2|=2\Rightarrow|\lambda^3+\lambda+2|=12$. Taking $\lambda^3+\lambda+2=12$ gives $\lambda^3+\lambda-10=0$, whose integer root is $\lambda=2$ (the other case $\lambda^3+\lambda+2=-12$ has no integer root).

    Then with $\lambda=2$:

    $|\lambda\hat i-3\lambda\hat j+3\hat k|=|2\hat i-6\hat j+3\hat k|=\sqrt{4+36+9}=\sqrt{49}=7.$

    ✓Final answer

    $7$ — option (C).