Q.State True or False: If $|\vec{a}|=|\vec{b}|$, then necessarily it implies $\vec{a}=\pm\vec{b}$.

Showing the 12 most recent of 21 on this concept.

1. 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:

   $$$$

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2. 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 $\cos\theta$. …

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3. 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 $|\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=1$:

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

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

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

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4. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $\vec{r}$ is a vector perpendicular to both the vectors $2\vec{i} + 3\vec{j} - 4\vec{k}$ and $3\vec{i} - \vec{j} + \vec{k}$ and satisfy $\vec{r} \cdot (3\vec{i} - 3\vec{j} + 4\vec{k}) = 5$ then $|\vec{r}| =$ (A) $\sqrt{366}$ (B) $\sqrt{222}$ (C) $\sqrt{318}$ (D) $\sqrt{246}$

   ›Reveal solutionSolution

   To find a vector perpendicular to two given vectors, we use their cross product. The scalar multiple of this cross product is then determined using the given dot product condition. Finally, we calculate the magnitude of the resulting vector. The magnitude of $\vec{r}$ is $\boxed{\sqrt{318}}$.

   When a vector is perpendicular to two other vectors, it means it lies along the direction of their cross product. The cross product of two vectors $\vec{a}$ and $\vec{b}$, denoted $\vec{a} \times \vec{b}$, results in a new vector that is perpendicular to both $\vec{a}$ and $\vec{b}$. Any vector perpendicular to both $\vec{a}$ and $\vec{b}$ must be a scalar multiple of $\vec{a} \times \vec{b}$.

   Once we establish this relationship, we can use the given dot product condition to find the specific scalar multiple. The dot product provides a way to relate the components of the unknown vector to a known scalar value, allowing us to solve for the scalar. Finally, with the complete vector determined, its magnitude can be calculated using the standard formula.

   Here's how to solve the problem step-by-step:

   1. Identify the given vectors and the condition:

      Let the two vectors be $\vec{a} = 2\vec{i} + 3\vec{j} - 4\vec{k}$ and $\vec{b} = 3\vec{i} - \vec{j} + \vec{k}$.

      We are looking for a vector $\vec{r}$ such that:

      • $\vec{r}$ is perpendicular to $\vec{a}$.
      • $\vec{r}$ is perpendicular to $\vec{b}$.
      • $\vec{r} \cdot (3\vec{i} - 3\vec{j} + 4\vec{k}) = 5$.

   2. Find a vector perpendicular to both $\vec{a}$ and $\vec{b}$:

      A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product, $\vec{a} \times \vec{b}$.

      The cross product of $\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}$ and $\vec{b} = b_1\vec{i} + b_2\vec{j} + b_3\vec{k}$ is:

      $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\vec{i} - (a_1b_3 - a_3b_1)\vec{j} + (a_1b_2 - a_2b_1)\vec{k}$

      Let's calculate $\vec{a} \times \vec{b}$:

   $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & -4 \\ 3 & -1 & 1 \end{vmatrix}$

   $= \vec{i}((3)(1) - (-4)(-1)) - \vec{j}((2)(1) - (-4)(3)) + \vec{k}((2)(-1) - (3)(3))$

   $= \vec{i}(3 - 4) - \vec{j}(2 - (-12)) + \vec{k}(-2 - 9)$

   $= \vec{i}(-1) - \vec{j}(14) + \vec{k}(-11)$

   $= -\vec{i} - 14\vec{j} - 11\vec{k}$

   3. Express $\vec{r}$ in terms of the cross product:

      Since $\vec{r}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, it must be parallel to $\vec{a} \times \vec{b}$. Therefore, $\vec{r}$ can be written as a scalar multiple of $\vec{a} \times \vec{b}$.

      Let $\vec{r} = \lambda (\vec{a} \times \vec{b})$ for some scalar $\lambda$.

      So, $\vec{r} = \lambda (-\vec{i} - 14\vec{j} - 11\vec{k})$.

   4. Use the dot product condition to find $\lambda$: …

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5. 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 $\overrightarrow{CB} = -\overrightarrow{BC} = -(3\hat{i} + 2\hat{j} - 2\hat{k}) = -3\hat{i} - 2\hat{j} + 2\hat{k}$. …
   

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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 $\lambda$,

   $$$$

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7. 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: $-2\vec{i} + \vec{j} + \vec{k} = 3\vec{a} - 3\vec{b}$. …

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8. 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:

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

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9. 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 $\vec c\cdot\vec b=1$: …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $x\vec{a} + y\vec{b} + z\vec{c} = p(\vec{b} \times \vec{c}) + q(\vec{c} \times \vec{a}) + r(\vec{a} \times \vec{b})$. If $(\vec{a}, \vec{b}) = (\vec{b}, \vec{c}) = (\vec{c}, \vec{a}) = \dfrac{\pi}{3}$, $(\vec{a}, \vec{b} \times \vec{c}) = \dfrac{\pi}{6}$ and $\vec{a}, \vec{b}, \vec{c}$ form a right-handed system, then $\dfrac{x+y+z}{p+q+r} =$ (A) $\dfrac{3}{4}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $2\sqrt{2}$ (D) $\dfrac{3}{8}$

    ›Reveal solutionSolution

    The key idea is to express the given vector equation in terms of a basis formed by $\vec{a}, \vec{b}, \vec{c}$ and use the given angles to compute dot products and scalar triple products, leading to $\frac{x+y+z}{p+q+r} = \frac{3}{8}$.

    We are given three unit vectors $\vec{a}, \vec{b}, \vec{c}$ with pairwise angles $\frac{\pi}{3}$, and the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $\frac{\pi}{6}$, with a right-handed system. The equation

    $x\vec{a} + y\vec{b} + z\vec{c} = p(\vec{b} \times \vec{c}) + q(\vec{c} \times \vec{a}) + r(\vec{a} \times \vec{b})$

    relates two linear combinations. The goal is to find $\frac{x+y+z}{p+q+r}$.

    Concept and intuition:

    Since $\vec{a}, \vec{b}, \vec{c}$ are not coplanar (they form a right-handed system and have a nonzero scalar triple product), they form a basis for 3D space. The right side uses cross products, which are perpendicular to the original vectors. To compare coefficients, we can take dot products with each of $\vec{a}, \vec{b}, \vec{c}$ to get equations linking $x,y,z$ to $p,q,r$. Then summing those equations yields the desired ratio.

    ---

    1. Compute the scalar triple product $[\vec{a}\,\vec{b}\,\vec{c}]$. For unit vectors with pairwise angles $\frac{\pi}{3}$, the volume of the parallelepiped is

    $[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}|\,|\vec{b} \times \vec{c}| \cos\frac{\pi}{6}.$

    Since $|\vec{b} \times \vec{c}| = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, we get

    $[\vec{a}\,\vec{b}\,\vec{c}] = 1 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4}.$

    This positive value confirms the right-handed system.

    2. Take dot product of the given equation with $\vec{a}$. Left side: $x(\vec{a}\cdot\vec{a}) + y(\vec{b}\cdot\vec{a}) + z(\vec{c}\cdot\vec{a}) = x + y\cos\frac{\pi}{3} + z\cos\frac{\pi}{3} = x + \frac{y}{2} + \frac{z}{2}$. Right side: $p(\vec{b}\times\vec{c})\cdot\vec{a} + q(\vec{c}\times\vec{a})\cdot\vec{a} + r(\vec{a}\times\vec{b})\cdot\vec{a}$.
       • $(\vec{b}\times\vec{c})\cdot\vec{a} = [\vec{a}\,\vec{b}\,\vec{c}] = \frac{3}{4}$.
       • $(\vec{c}\times\vec{a})\cdot\vec{a} = 0$ (cross product perpendicular to $\vec{a}$).
       • $(\vec{a}\times\vec{b})\cdot\vec{a} = 0$ (same reason). So right side = $p \cdot \frac{3}{4}$. Equation (1):

    $x + \frac{y}{2} + \frac{z}{2} = \frac{3}{4}p.$

    3. Take dot product with $\vec{b}$. Left: $x\cos\frac{\pi}{3} + y + z\cos\frac{\pi}{3} = \frac{x}{2} + y + \frac{z}{2}$. Right: $p(\vec{b}\times\vec{c})\cdot\vec{b} + q(\vec{c}\times\vec{a})\cdot\vec{b} + r(\vec{a}\times\vec{b})\cdot\vec{b}$.
       • $(\vec{b}\times\vec{c})\cdot\vec{b} = 0$.
       • $(\vec{c}\times\vec{a})\cdot\vec{b} = [\vec{b}\,\vec{c}\,\vec{a}] = [\vec{a}\,\vec{b}\,\vec{c}] = \frac{3}{4}$ (cyclic permutation).
       • $(\vec{a}\times\vec{b})\cdot\vec{b} = 0$. So right side = $q \cdot \frac{3}{4}$. Equation (2):

    $\frac{x}{2} + y + \frac{z}{2} = \frac{3}{4}q.$

    4. Take dot product with $\vec{c}$. …

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11. 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:

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

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.A, B, C, D are any four points. If E and F are mid points of AC and BD respectively, then $\overline{AB} + \overline{CB} + \overline{CD} + \overline{AD} =$ (A) $\overline{EF}$ (B) $2\overline{EF}$ (C) $3\overline{EF}$ (D) $4\overline{EF}$

    ›Reveal solutionSolution

    The sum of the four vectors equals four times the vector joining the midpoints of the diagonals, so the answer is 4 EF.

    ---

    Concept and Intuition

    We are dealing with vectors in free space — points A, B, C, D are arbitrary, and we want a neat expression for

    $\overrightarrow{AB} + \overrightarrow{CB} + \overrightarrow{CD} + \overrightarrow{AD}$

    in terms of the segment joining the midpoints of the diagonals AC and BD.

    The key trick: midpoint vectors let us rewrite each side as a difference of position vectors. Then, by grouping terms cleverly, the sum collapses into a multiple of $\overrightarrow{EF}$.

    ---

    Step‑by‑Step Solution

    1. Assign position vectors Let the position vectors of A, B, C, D be $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ respectively (with respect to some origin). Then:

    $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \quad \overrightarrow{CB} = \mathbf{b} - \mathbf{c}, \quad \overrightarrow{CD} = \mathbf{d} - \mathbf{c}, \quad \overrightarrow{AD} = \mathbf{d} - \mathbf{a}.$

    2. Write the sum

    $S = (\mathbf{b} - \mathbf{a}) + (\mathbf{b} - \mathbf{c}) + (\mathbf{d} - \mathbf{c}) + (\mathbf{d} - \mathbf{a}).$

    3. Collect like terms

       • Terms with $\mathbf{b}$: $\mathbf{b} + \mathbf{b} = 2\mathbf{b}$
       • Terms with $\mathbf{d}$: $\mathbf{d} + \mathbf{d} = 2\mathbf{d}$
       • Terms with $-\mathbf{a}$: $-\mathbf{a} - \mathbf{a} = -2\mathbf{a}$
       • Terms with $-\mathbf{c}$: $-\mathbf{c} - \mathbf{c} = -2\mathbf{c}$

       So

    $S = 2\mathbf{b} + 2\mathbf{d} - 2\mathbf{a} - 2\mathbf{c} = 2\bigl[(\mathbf{b} + \mathbf{d}) - (\mathbf{a} + \mathbf{c})\bigr].$

    4. Introduce midpoints E is the midpoint of AC, so $\mathbf{e} = \frac{\mathbf{a} + \mathbf{c}}{2}.$ …

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