Question

Let \(\vec{a}\) be a vector which is perpendicular to the vector 
\(3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}. \)If \(\vec{a}×(2\hat{i}+\hat{k})=2\hat{i}−13\hat{j}−4\hat{k}\)
, then the projection of the vector on the vector
\( 2\hat{i}+2\hat{j}+\hat{k}\) is:

• A. \(\frac{1}{3}\)
• B. 1
• C. \(\frac{5}{3}\)
• D. \(\frac{7}{3}\)

Answer 1

The Correct Option is C

The correct answer is (C) : \(\frac{5}{3}\)
Let \(\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\)
and
\(\vec{a}⋅(3\hat{i}−\frac{1}{2}\hat{j}+2\hat{k})=0⇒3a_1+\frac{a_2}{2}+2a_3=0…(i)\)
and
\(\vec{a}×(2\hat{i}+\hat{k})=2\hat{i}−13\hat{j}−4\hat{k}\)
\(⇒a_2\hat{i}+(2a_3−a_1)\hat{j}−2a_2\hat{k}=2\hat{i}−13\hat{j}−4\hat{k}\)
∴ a₂ = 2 …(ii)
and a₁ – 2a₃ = 13 …(iii)
From eq. (i) and (iii) : a₁ = 3 and a₃ = –5
\(∴\vec{a}=3\hat{i}+2\hat{j}−5\hat{k}\)
∴projection of \(\vec{a}\) on \(2\hat{i}+2\hat{j}+\hat{k}=\frac{6+4−5}{3}=\frac{5}{3}\)