Question

If \(\vec{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{c}=\hat{i}-\hat{j}\) are such that \(\vec{a}+μ\vec{b}\) is perpendicular to \(\vec{c}\), then the value of μ.

(a) \(\frac{7}{2}\)

(b) –\(\frac{7}{2}\)

(c) –\(\frac{3}{2}\)

(d) \(\frac{7}{9}\)

This question was addressed to me during an interview.

This intriguing question originated from Product of Two Vectors-2 topic in portion Vector Algebra of Mathematics – Class 12

Answer 1

Right choice is (b) –\(\frac{7}{2}\)

Explanation: Given that: \(\vec{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{c}=\hat{i}-\hat{j}\)

Also given, \(\vec{a}+μ\vec{b}\) is perpendicular to \(\vec{c}\)

Therefore, \((\vec{a}+μ\vec{b}).\vec{c}=0\)

i.e. \((\hat{i}-\hat{j}+3\hat{k}+μ(5\hat{i}-2\hat{j}+\hat{k})).(\hat{i}-\hat{j})\)=0

\(((1+5μ) \,\hat{i}-(1+2μ) \,\hat{j}+(μ+3) \,\hat{k}).(\hat{i}-\hat{j})\)=0

1+5μ+1+2μ=0

μ=-\(\frac{7}{2}\).