Question

Let $\vec a=\hat i+\hat j−\hat k$ and $\vec c=2\hat i−3\hat j+2\hat k$. Then the number of vectors $\vec b$ such that $\vec b×\vec c=\vec a$ and $|\vec b|∈{1,2,…,10}$ is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

$\vec a=\hat i+\hat j−\hat k$
$\vec c=2\hat i−3\hat j+2\hat k$
Then, $\vec b×\vec c=\vec a$
$⇒ \vec c.(\vec b \times \vec c) = \vec c. \vec a$
$\vec c. \vec a = 0$
$⇒(\hat i+\hat j−\hat k)(2\hat i−3\hat j+2\hat k) = 0$
$⇒ 2 – 3 – 2 = 0$
$⇒$ $–3 = 0$ (possibility null)
⇒ No value of $\vec b$ is possible.

Therefore, the correct option is (A): $0$