Question

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

• A. 193
• B. 218
• C. 205
• D. 233

Hint:

If $(\vec{u}\times\vec{v})=\vec{0}$, then $\vec{u}$ and $\vec{v}$ are parallel — use this to reduce vector equations quickly.

Answer 1

The Correct Option is B

Step 1: Use the vector identity.
Given:
2(a × c) + 3(b × c) = 0

Factoring c:
(2a + 3b) × c = 0

Hence, vector c is parallel to (2a + 3b).

Step 2: Compute 2a + 3b.
2a = 4i − 10j + 10k
3b = 3i − 3j + 9k

2a + 3b = 7i − 13j + 19k

Thus,
c = λ(7i − 13j + 19k)

Step 3: Use the dot product condition.
a − b = (2 − 1)i + (−5 + 1)j + (5 − 3)k
a − b = i − 4j + 2k

(a − b) · c = λ(7 + 52 + 38)
(a − b) · c = 97λ

Given:
97λ = −97
λ = −1

Step 4: Find c × k.
c = −7i + 13j − 19k

c × k = 13i + 7j

Step 5: Compute the required value.
|c × k|² = 13² + 7²
|c × k|² = 169 + 49
|c × k|² = 218