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Mathematics Question on Vectors

Let a⃗\vec{a} be a vector which is perpendicular to the vector
3i^+12j^+2k^.3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}. If a⃗×(2i^+k^)=2i^−13j^−4k^\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
2i^+2j^+k^2\hat{i}+2\hat{j}+\hat{k} is:

A

13\frac{1}{3}

B

1

C

53\frac{5}{3}

D

73\frac{7}{3}

Answer

53\frac{5}{3}

Explanation

Solution

The correct answer is (C) : 53\frac{5}{3}
Let a⃗=a1i^+a2j^+a3k^\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}
and
a⃗⋅(3i^−12j^+2k^)=0⇒3a1+a22+2a3=0…(i)\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
a⃗×(2i^+k^)=2i^−13j^−4k^\vec{a}×(2\hat{i}+\hat{k})=2\hat{i}−13\hat{j}−4\hat{k}
⇒a2i^+(2a3−a1)j^−2a2k^=2i^−13j^−4k^⇒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
∴a⃗=3i^+2j^−5k^∴\vec{a}=3\hat{i}+2\hat{j}−5\hat{k}
∴projection of a⃗\vec{a} on 2i^+2j^+k^=6+4−53=532\hat{i}+2\hat{j}+\hat{k}=\frac{6+4−5}{3}=\frac{5}{3}