Let vec a =2 hat i -5 hat j +5 k and vec b =hat i -hat j +3 k . If vec c is a vector such that 2(vec a × vec c )+3(vec b × vec c )=overrightarrow 0 and (vec a -vec b ) × vec c =-97 , then overrightarrow c × k 2 is equal to

Mathematics · Vector Algebra

JEE Main 2026 — 24 January, Evening Shift — Question 19

Let $\vec{a}=2 \hat{i}-5 \hat{j}+5 k$ and $\vec{b}=\hat{i}-\hat{j}+3 k$ . If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ and $(\vec{a}-\vec{b}) \cdot \vec{c}=-97$ , then $|\overrightarrow{\mathrm{c}} \times \mathrm{k}|^{2}$ is equal to

Option A:
193
Option B:
233
Option C:
218
Correct
Option D:
205

Answer: C

Step-by-step solution

\vec a=2\hat i-5\hat j+5\hat k,\qquad \vec b=\hat i-\hat j+3\hat k

Given

2(\vec a\times \vec c)+3(\vec b\times \vec c)=\vec 0

Using distributivity of cross product,

(2\vec a+3\vec b)\times \vec c=\vec 0

Hence $\vec c$ is parallel to $2\vec a+3\vec b$ .

Compute:

2\vec a+3\vec b =2(2,-5,5)+3(1,-1,3) =(7,-13,19)

So let

\vec c=\lambda(7,-13,19)

Now use the dot product condition:

(\vec a-\vec b)\cdot \vec c=-97

First,

\vec a-\vec b=(1,-4,2)

Hence

(1,-4,2)\cdot \lambda(7,-13,19) =\lambda(7+52+38) =97\lambda

So,

97\lambda=-97 \;\Rightarrow\; \lambda=-1

Thus,

\vec c=(-7,13,-19)

Now,

\vec c\times \hat k= \begin{vmatrix} \hat i & \hat j & \hat k\\ -7 & 13 & -19\\ 0 & 0 & 1 \end{vmatrix} =13\hat i+7\hat j

Therefore,

|\vec c\times \hat k|^2 =13^2+7^2 =169+49 =218

\boxed{218}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Vector or Cross Product of Two Vectors

Answer: C

Step-by-step solution

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