Question

Let \(\vec{a}=\hat{i}-\hat{j}\), \(\vec{b}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{c}\) be a vector such that \(\vec{a}\times\vec{c}+\vec{b}=\vec{0}\) and \(\vec{a}\cdot\vec{c}=4\), then \(|\vec{c}|^2\) is equal to :

• A. 8
• B. \(\frac{19}{2}\)
• C. 9
• D. \(\frac{17}{2}\)

Hint:

Write \(\vec c=(x,y,z)\), expand \(\vec a\times\vec c=-\vec b\) in components and combine with \(\vec a\cdot\vec c=4\).

Answer 1

The Correct Option is B

Let \(\vec c=x\hat i+y\hat j+z\hat k\).
\(\vec a\times\vec c=(1,-1,0)\times(x,y,z)=(-z,\,-z,\,x+y)\).
\(\vec a\times\vec c=-\vec b=(-1,-1,-1)\Rightarrow z=1,\;x+y=-1\).
\(\vec a\cdot\vec c=x-y=4\).
Solving: \(x=\frac32,\;y=-\frac52\).
\(|\vec c|^2=\frac94+\frac{25}{4}+1=\frac{38}{4}=\frac{19}{2}\).
Answer: (2) \(\frac{19}{2}\)