Test Index

Vector Algebra Part 3

Section: Mathematics

© examsnet.com

Question : 6 of 72

Marks: +1, -0

Let

\vec{a}=2\hat{i}-5\hat{j}+5\hat{k}

\vec{b}=\hat{i}-\hat{j}+3\hat{k}

\vec{c}

is a vector such that

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

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

|\vec{c}\times\hat{k}|^2

[24 Jan 2026 Shift 2]

193
205
218
233

Solution:

Concept:

The given vector equation implies

2\vec{a}+3\vec{b}

is parallel to

\vec{c}

.

Explanation:

From

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

, we have

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

.

Thus

\vec{c}=\lambda(2\vec{a}+3\vec{b})

for some scalar

\lambda

.

Compute

2\vec{a}+3\vec{b}

:

2(2\hat{i}-5\hat{j}+5\hat{k})+3(\hat{i}-\hat{j}+3\hat{k}) = 7\hat{i}-13\hat{j}+19\hat{k}

.

So

\vec{c}=\lambda(7\hat{i}-13\hat{j}+19\hat{k})

.

Now

\vec{a}-\vec{b} = (2\hat{i}-5\hat{j}+5\hat{k})-(\hat{i}-\hat{j}+3\hat{k}) = \hat{i}-4\hat{j}+2\hat{k}

.

Given

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

, substitute:

(\hat{i}-4\hat{j}+2\hat{k})\cdot\lambda(7\hat{i}-13\hat{j}+19\hat{k}) = \lambda(7+52+38) = 97\lambda = -97

.

Hence

\lambda = -1

and

\vec{c} = -7\hat{i}+13\hat{j}-19\hat{k}

.

Compute

\vec{c}\times\hat{k}

:

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

.

Magnitude squared:

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

.

Answer:

C. 218

© examsnet.com

Go to Question:

Prev Question

Next Question