Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}$ and $\vec{c}=\hat{j}-\hat{k}$ be three vectors such that $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$. If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $\ell$, then the value of $3 \ell^{2}$ is equal to ____.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Vector Algebra

Let

a = \hat{i} + \hat{j} + \hat{k}, b

and

c = \hat{j} - \hat{k}

be three vectors such that

a \times b = c

and

a \cdot b = 1

. If the length of projection vector of the vector

b

on the vector

a \times c

ℓ

, then the value of

3 ℓ^{2}

is equal to ____.

Answer: 2

Given

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

c = \hat{j} - \hat{k}

a \times b = c

, and

a \cdot b = 1

.

First, find

a \times c

a \times c = \hat{i} 10 \hat{j} 11 \hat{k} 1 - 1 = \hat{i} (- 1 - 1) - \hat{j} (- 1 - 0) + \hat{k} (1 - 0) = - 2 \hat{i} + \hat{j} + \hat{k}

∣ a \times c ∣ = 4 + 1 + 1 = 6

.

The projection length is:

ℓ = \frac{∣ b \cdot ( a \times c ) ∣}{∣ a \times c ∣}

Using the scalar triple product property:

b \cdot (a \times c) = c \cdot (b \times a) = - c \cdot (a \times b) = - c \cdot c = - ∣ c ∣^{2} = - (1 + 1) = - 2

.

Therefore:

ℓ = \frac{∣ - 2∣}{6} = \frac{2}{6} = \frac{6}{3}

3 ℓ^{2} = 3 \cdot \frac{6}{9} = 3 \cdot \frac{2}{3} = 2

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer