Q.If $\begin{bmatrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+2 \\ 2b+4 & -21 & 0 \end{bmatrix}$ Find the values of $a, b, c, x, y$ and $z$.

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If $\left(\dfrac{9}{4}, \dfrac{5}{4}, \dfrac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1)$, $(1, b, 4)$, $(4, 0, c)$ and $(1, 1, 7)$, then (A) $a = b = c$ (B) $a = b = c + 1$ (C) $b = c = a + 1$ (D) $a = c = b + 1$

   ›Reveal solutionSolution

   The centroid of a tetrahedron is the average of its four vertices. Equating the given centroid to the average yields three equations, which give $a=2$, $b=3$, $c=4$, so $b = c = a + 1$.

   The centroid of a tetrahedron is not the same as the centroid of a triangle — but the idea is similar. For a triangle, the centroid is the average of the three vertices. For a tetrahedron (a 3D solid with four vertices), the centroid is simply the average of the coordinates of its four vertices. That’s the key concept: if the vertices are $A$, $B$, $C$, $D$, then the centroid $G$ is

   $G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4}, \frac{z_A + z_B + z_C + z_D}{4} \right).$

   We are told this centroid equals $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$. So we just match coordinates and solve.

   1. Set up the x-coordinate equation. The x-coordinates of the vertices are $a$, $1$, $4$, $1$. Their sum is $a + 1 + 4 + 1 = a + 6$. The centroid’s x-coordinate is $\frac{a+6}{4}$, and this must equal $\frac{9}{4}$.

   $\frac{a+6}{4} = \frac{9}{4} \quad\Rightarrow\quad a+6 = 9 \quad\Rightarrow\quad a = 3.$

   2. Set up the y-coordinate equation. The y-coordinates are $2$, $b$, $0$, $1$. Their sum is $2 + b + 0 + 1 = b + 3$. The centroid’s y-coordinate is $\frac{b+3}{4}$, and this must equal $\frac{5}{4}$.

   $\frac{b+3}{4} = \frac{5}{4} \quad\Rightarrow\quad b+3 = 5 \quad\Rightarrow\quad b = 2.$

   3. Set up the z-coordinate equation. The z-coordinates are $1$, $4$, $c$, $7$. Their sum is $1 + 4 + c + 7 = c + 12$. The centroid’s z-coordinate is $\frac{c+12}{4}$, and this must equal $\frac{15}{4}$.

   $\frac{c+12}{4} = \frac{15}{4} \quad\Rightarrow\quad c+12 = 15 \quad\Rightarrow\quad c = 3.$

   So we have $a = 3$, $b = 2$, $c = 3$. Now check the options:

   • (A) $a = b = c$? No, $3 = 2$ is false.
   • (B) $a = b = c + 1$? $3 = 2$ is false.
   • (C) $b = c = a + 1$? $2 = 3$ is false, and $3 = 3+1$ is false.
   • (D) $a = c = b + 1$? $3 = 3$ is true, and $3 = 2+1$ is true.

   Watch out

   A common mistake is to treat the centroid as the average of three vertices (like a triangle) or to forget to divide by 4. Always use 4 for a tetrahedron.

   ✓Final answer

   The correct option is (D).

2. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}$, $\vec{b} = 3(\vec{i} - \vec{j} + \vec{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c}) =$ (A) $32$ (B) $24$ (C) $20$ (D) $36$

   ›Reveal solutionSolution

   The key idea is to use vector identities to simplify the expression $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c})$ into a form involving known dot and cross products, then substitute the given values to get the result $24$, which corresponds to option (B).

   We are given:

   $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}, \quad \vec{b} = 3(\vec{i} - \vec{j} + \vec{k}), \quad \vec{a} \times \vec{c} = \vec{b}, \quad \vec{a} \cdot \vec{c} = 3.$

   We need $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c})$.

   Concept and intuition:

   The expression mixes dot and cross products. The term $\vec{c} \times \vec{b}$ is perpendicular to both $\vec{c}$ and $\vec{b}$, but when dotted with $\vec{a}$, we can use the scalar triple product identity: $\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{c} \cdot (\vec{b} \times \vec{a})$. Since we know $\vec{a} \times \vec{c} = \vec{b}$, we can relate $\vec{b} \times \vec{a}$ to something simpler. The other terms $\vec{a} \cdot \vec{b}$ and $\vec{a} \cdot \vec{c}$ are directly computable or given. This avoids solving for $\vec{c}$ explicitly.

   Step-by-step solution:

   1. Simplify the triple product term. Use the scalar triple product cyclic property:

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

   Now, $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. But we know $\vec{a} \times \vec{c} = \vec{b}$. To relate $\vec{a} \times \vec{b}$, take the cross product of both sides of $\vec{a} \times \vec{c} = \vec{b}$ with $\vec{a}$:

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

   Use the vector triple product identity: $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.

   So:

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

   Given $\vec{a} \cdot \vec{c} = 3$, and $|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6$, we have:

   $\vec{a} \times \vec{b} = 3\vec{a} - 6\vec{c}.$

   Hence:

   $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) = -3\vec{a} + 6\vec{c}.$

   Therefore:

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{c} \cdot (\vec{b} \times \vec{a}) = \vec{c} \cdot (-3\vec{a} + 6\vec{c}) = -3(\vec{a} \cdot \vec{c}) + 6|\vec{c}|^2.$

   Since $\vec{a} \cdot \vec{c} = 3$, this becomes:

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = -9 + 6|\vec{c}|^2.$

   2. Find $|\vec{c}|^2$ using the given cross product. From $\vec{a} \times \vec{c} = \vec{b}$, take the magnitude squared:

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

   We know $|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2 |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$ (Lagrange's identity).

   Compute $|\vec{b}|^2$: $\vec{b} = 3(1, -1, 1)$, so $|\vec{b}|^2 = 9(1+1+1) = 27$.

   Also $|\vec{a}|^2 = 6$, $\vec{a} \cdot \vec{c} = 3$.

   Thus:

   $6|\vec{c}|^2 - 9 = 27 \quad \Rightarrow \quad 6|\vec{c}|^2 = 36 \quad \Rightarrow \quad |\vec{c}|^2 = 6.$

   3. Plug back into the triple product term.

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = -9 + 6 \times 6 = -9 + 36 = 27.$

   4. Compute the remaining terms. The full expression is:

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

   We already have $\vec{a} \cdot (\vec{c} \times \vec{b}) = 27$ and $\vec{a} \cdot \vec{c} = 3$.

   Now compute $\vec{a} \cdot \vec{b}$:

   $\vec{a} \cdot \vec{b} = (1,2,1) \cdot (3, -3, 3) = 3 - 6 + 3 = 0.$

   So:

   $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c}) = 27 - 0 - 3 = 24.$

   Tip

   Notice that $\vec{a} \cdot \vec{b} = 0$ tells us $\vec{a}$ and $\vec{b}$ are perpendicular — a quick check that saves arithmetic.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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

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

   5. Add the three equations. Left sum: $(x + \frac{y}{2} + \frac{z}{2}) + (\frac{x}{2} + y + \frac{z}{2}) + (\frac{x}{2} + \frac{y}{2} + z)$ = $(x + \frac{x}{2} + \frac{x}{2}) + (\frac{y}{2} + y + \frac{y}{2}) + (\frac{z}{2} + \frac{z}{2} + z)$ = $(2x) + (2y) + (2z) = 2(x+y+z)$. Right sum: $\frac{3}{4}(p+q+r)$. Hence

   $2(x+y+z) = \frac{3}{4}(p+q+r).$

   6. Solve for the ratio.

   $\frac{x+y+z}{p+q+r} = \frac{3}{8}.$

   Watch out

   A common mistake is to forget that $(\vec{c}\times\vec{a})\cdot\vec{b}$ equals the scalar triple product $[\vec{b}\,\vec{c}\,\vec{a}]$, which is the same as $[\vec{a}\,\vec{b}\,\vec{c}]$ for a cyclic permutation, not zero.

   Tip

   The symmetry of the problem makes summing the three dot-product equations much faster than solving for individual variables.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. 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

5. 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

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

7. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Let $\vec{a} = \lambda \hat{i} + 3\hat{j} + 4\hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} + \lambda \hat{k}$ and $\vec{c} = \lambda \hat{i} + \hat{j} - 3\hat{k}$ be three vectors for some integer $\lambda$. If the volume of the parallelepiped with $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is $61$ cubic units, then the number of possible values of $\lambda$ is (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   The volume of a parallelepiped is the absolute value of the scalar triple product. Setting $|[\vec{a},\vec{b},\vec{c}]| = 61$ gives a cubic equation in $\lambda$; counting its integer solutions yields the answer.

   Concept & Intuition

   The volume of a parallelepiped formed by three vectors is the absolute value of their scalar triple product:

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

   This is also the absolute value of the determinant of the matrix whose rows (or columns) are the vectors.

   We are told $V = 61$, so we compute the determinant, set its absolute value equal to 61, and solve for the integer $\lambda$. The number of integer solutions is what we count.

   Step-by-step solution

   1. Write the vectors in component form

   $\vec{a} = (\lambda,\;3,\;4),\quad \vec{b} = (3,\;-1,\;\lambda),\quad \vec{c} = (\lambda,\;1,\;-3).$

   2. Compute the scalar triple product The scalar triple product $[\vec{a},\vec{b},\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$ equals the determinant:

   $\Delta = \begin{vmatrix} \lambda & 3 & 4 \\ 3 & -1 & \lambda \\ \lambda & 1 & -3 \end{vmatrix}.$

   Expand along the first row:

   $\Delta = \lambda \begin{vmatrix} -1 & \lambda \\ 1 & -3 \end{vmatrix} - 3 \begin{vmatrix} 3 & \lambda \\ \lambda & -3 \end{vmatrix} + 4 \begin{vmatrix} 3 & -1 \\ \lambda & 1 \end{vmatrix}.$

   Compute each minor:

   $\begin{vmatrix} -1 & \lambda \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (\lambda)(1) = 3 - \lambda,$

   $\begin{vmatrix} 3 & \lambda \\ \lambda & -3 \end{vmatrix} = (3)(-3) - (\lambda)(\lambda) = -9 - \lambda^2,$

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

   Substitute back:

   $\Delta = \lambda(3 - \lambda) - 3(-9 - \lambda^2) + 4(3 + \lambda).$

   3. Simplify the expression

   $\Delta = 3\lambda - \lambda^2 + 27 + 3\lambda^2 + 12 + 4\lambda.$

   Combine like terms:

   $\Delta = (-\lambda^2 + 3\lambda^2) + (3\lambda + 4\lambda) + (27 + 12) = 2\lambda^2 + 7\lambda + 39.$

   4. Set the absolute value equal to 61 Volume is $|\Delta| = 61$, so:

   $|2\lambda^2 + 7\lambda + 39| = 61.$

   This gives two cases:

   $2\lambda^2 + 7\lambda + 39 = 61 \quad \text{or} \quad 2\lambda^2 + 7\lambda + 39 = -61.$

   5. Solve each equation
      • First equation:

   $2\lambda^2 + 7\lambda + 39 = 61 \implies 2\lambda^2 + 7\lambda - 22 = 0.$

    Discriminant: $7^2 - 4(2)(-22) = 49 + 176 = 225 = 15^2$.  
    Roots:  
   

   $\lambda = \frac{-7 \pm 15}{4} \implies \lambda = 2 \quad \text{or} \quad \lambda = -\frac{11}{2}.$

    Only $\lambda = 2$ is an integer.
   

   • Second equation:

   $2\lambda^2 + 7\lambda + 39 = -61 \implies 2\lambda^2 + 7\lambda + 100 = 0.$

    Discriminant: $7^2 - 4(2)(100) = 49 - 800 = -751 < 0$.  
    No real solutions, hence no integer solutions.
   

   6. Count the integer values

   Only $\lambda = 2$ works. So there is exactly 1 possible integer value.

   Watch out

   A common mistake is to forget the absolute value and solve only $2\lambda^2+7\lambda+39=61$, missing that the volume is always positive. Here the other case gave no real roots, but in other problems it might yield extra solutions.

   Tip

   The discriminant $225$ is a perfect square, so the quadratic factors nicely: $2\lambda^2+7\lambda-22 = (2\lambda+11)(\lambda-2)$. This confirms the integer root $\lambda=2$ immediately.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. 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 $\boxed{\dfrac{\sqrt{15}}{8}}$.

9. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are the position vectors of three points A, B, C respectively. If $\angle ABC = \dfrac{\pi}{2}$, $\overrightarrow{AB} = \vec{i} + 4\vec{j} + (4 - \lambda)\vec{k}$, $\overrightarrow{AC} = (\lambda - 1)\vec{i} + 6\vec{j} + (2 - \lambda)\vec{k}$, then $\lambda =$ (A) $0$ (B) $-\dfrac{1}{3}$ (C) $-\dfrac{3}{4}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   The right angle at $B$ forces $\lambda = \dfrac{2}{3}$ (D).

   $\angle ABC=\dfrac{\pi}{2}$ means $\overrightarrow{BA}\perp\overrightarrow{BC}$. Writing these through the given vectors, $\overrightarrow{BA}=-\overrightarrow{AB}$ and $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$, so

   $\overrightarrow{BA}\cdot\overrightarrow{BC}=-\overrightarrow{AB}\cdot(\overrightarrow{AC}-\overrightarrow{AB})=|\overrightarrow{AB}|^2-\overrightarrow{AB}\cdot\overrightarrow{AC}=0\ \Rightarrow\ |\overrightarrow{AB}|^2=\overrightarrow{AB}\cdot\overrightarrow{AC}.$

   With $\overrightarrow{AB}=\vec i+4\vec j+(4-\lambda)\vec k$ and $\overrightarrow{AC}=(\lambda-1)\vec i+6\vec j+(2-\lambda)\vec k$:

   $|\overrightarrow{AB}|^2 = 1+16+(4-\lambda)^2=\lambda^2-8\lambda+33,$

   $\overrightarrow{AB}\cdot\overrightarrow{AC}=(\lambda-1)+24+(4-\lambda)(2-\lambda)=\lambda^2-5\lambda+31.$

   Setting them equal:

   $\lambda^2-8\lambda+33=\lambda^2-5\lambda+31\ \Rightarrow\ -3\lambda=-2\ \Rightarrow\ \lambda=\frac{2}{3}.$

   ✓Final answer

   $\lambda = \dfrac{2}{3}$ — option (D).

10. 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).

11. 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).

12. 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).