Question

Let \(\vec{u}\) be a vector coplanar with the vectors \(\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\) and \(\vec{b}=\hat{j}+\hat{k}\). If \(\vec{u}\) is perpendicular to \(\vec{a}\) and \(\vec{u}\cdot\vec{b}=24\), then \(|\vec{u}|^2\) is equal to

• A. 336
• B. 315
• C. 256
• D. 84

Hint:

Write \(\vec u=\lambda\vec a+\mu\vec b\) (coplanar) and use \(\vec u\cdot\vec a=0\), \(\vec u\cdot\vec b=24\).

Answer 1

The Correct Option is A

Since \(\vec u\) is coplanar with \(\vec a,\vec b\): \(\vec u=\lambda\vec a+\mu\vec b\). Here \(|\vec a|^2=14\), \(|\vec b|^2=2\), \(\vec a\cdot\vec b=3-1=2\).
\(\vec u\cdot\vec a=0\Rightarrow14\lambda+2\mu=0\Rightarrow\mu=-7\lambda\).
\(\vec u\cdot\vec b=24\Rightarrow2\lambda+2\mu=24\Rightarrow\lambda+\mu=12\).
So \(-6\lambda=12\Rightarrow\lambda=-2,\ \mu=14\).
\(\vec u=-2(2,3,-1)+14(0,1,1)=(-4,8,16)\).
\(|\vec u|^2=16+64+256=336\).
Answer: (1) 336