Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

(C) $\frac{2}{3} \sqrt{\frac{7}{3}}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Vector Algebra

Let

a = \hat{i} - \hat{j} + 2 \hat{k}

and

b

be a vector such that

a \times b = 2 \hat{i} - \hat{k}

and

a \cdot b = 3

. Then the projection of

b

on the vector

a - b

is :-

Answer: (A)

Given

a = (1, - 1, 2)

a \times b = (2, 0, - 1)

a \cdot b = 3

.

From

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}

4 + 0 + 1 = 6∣ b ∣^{2} - 9

, so

5 = 6∣ b ∣^{2} - 9

, giving

∣ b ∣^{2} = \frac{14}{6} = \frac{7}{3}

.

Projection of

b

a - b

\frac{b \cdot ( a - b )}{∣ a - b ∣}

b \cdot (a - b) = a \cdot b - ∣ b ∣^{2} = 3 - \frac{7}{3} = \frac{2}{3}

∣ a - b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2} = 6 - 6 + \frac{7}{3} = \frac{7}{3}

, so

∣ a - b ∣ = \frac{7}{3}

.

Projection =

\frac{2/3}{7/3} = \frac{2}{3} \cdot \frac{3}{7} = \frac{2}{21}

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option