Let \veca=\hati+2 \hatj+\hatk, \vecb=3 \hati-3 \hatj+3 \hatk, \vecc=2 \hati-\hatj+2 \hatk and \vecd be a vector such that \vecb \times \vecd=\vecc \times \vecd and \veca \cdot \vecd=4 . Then |(\veca \times \vecd)|^2 is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2025 · 3 Apr · Shift 2 · Q49 with Solution

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Correct answer: 128

$\vec a=(1,2,1),\quad \vec b=(3,-3,3),\quad \vec c=(2,-1,2)$

and a vector $\vec d$ such that

$\vec b\times \vec d=\vec c\times \vec d$

and

$\vec a\cdot \vec d=4.$

We need to find:

$|\vec a\times \vec d|^2.$

From

$\vec b\times \vec d=\vec c\times \vec d,$

we get

$(\vec b-\vec c)\times \vec d=\vec 0.$

Now,

$\vec b-\vec c=(3-2,-3-(-1),3-2)=(1,-2,1).$

So,

$(1,-2,1)\times \vec d=\vec 0.$

Hence $\vec d$ is parallel to $(1,-2,1)$ . Therefore let

$\vec d=\lambda(1,-2,1).$

Given

$\vec a\cdot \vec d=4,$

$(1,2,1)\cdot \lambda(1,-2,1)=4.$

Compute the dot product:

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

Thus,

$\vec d=-2(1,-2,1)=(-2,4,-2).$

Since

$\vec d=-2(1,-2,1),$

notice that

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

but we directly compute the cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 1\\ -2 & 4 & -2 \end{vmatrix}.$$ Expanding, $$\vec a\times \vec d= \hat i(2\cdot (-2)-1\cdot 4) -\hat j(1\cdot (-2)-1\cdot (-2)) +\hat k(1\cdot 4-2\cdot (-2)).$$ $$=\hat i(-4-4)-\hat j(-2+2)+\hat k(4+4)$$ $$=-8\hat i+0\hat j+8\hat k.$$ So, $$\vec a\times \vec d=(-8,0,8).$$ --- 5. **Find its squared magnitude** $$|\vec a\times \vec d|^2=(-8)^2+0^2+8^2=64+64=128.$$ --- 6. **Final answer** $$|\vec a\times \vec d|^2=128.$$ This matches the stored correct answer.