Question

Let \(\vec{x}\) be a vector in the plane containing vectors
\(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\)
and
\(\vec{b} = \hat{i} + 2\hat{j} - \hat{k}\).
If the vector \(\vec{x}\) is perpendicular to \(\left(3\hat{i} + 2\hat{j} - \hat{k}\right)\) and its projection on \(\vec{a}\) is \(\frac{17\sqrt{6}}{2}\), then the value of \(|\vec{x}|^2\) is equal to __________.

Hint:

Write \(\vec x = \lambda\vec a + \mu\vec b\), use the perpendicularity condition to fix the ratio \(\lambda:\mu\), then the projection fixes the scale.

Answer 1

Correct Answer: 486

Let \(\vec x = \lambda\vec a + \mu\vec b = (2\lambda+\mu,\ -\lambda+2\mu,\ \lambda-\mu)\).
\(\vec x\cdot(3,2,-1) = 6\lambda+3\mu - 2\lambda+4\mu - \lambda+\mu = 3\lambda + 8\mu = 0\), so \(\lambda = -\frac{8\mu}{3}\).
Take \(\mu = 3k\), \(\lambda = -8k\): \(\vec x = k(-13, 14, -11)\).
\(|\vec a| = \sqrt6\) and \(\vec x\cdot\vec a = k(-26 - 14 - 11) = -51k\).
Projection \(= \frac{|\vec x\cdot\vec a|}{|\vec a|} = \frac{51|k|}{\sqrt6} = \frac{17\sqrt6}{2}\), so \(|k| = \frac{17\cdot6}{2\cdot51} = 1\).
\(|\vec x|^2 = k^2(169 + 196 + 121) = 486\).
Answer: 486