Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$, whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$, is given by

(C) $3 \hat{i}-\hat{j}+3 \hat{k}$

(A) $\hat{i}-3 \hat{j}+3 \hat{k}$

(B) $-3 \hat{i}-3 \hat{j}-\hat{k}$

(D) $\hat{i}+3 \hat{j}-3 \hat{k}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Vector Algebra

Let

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

and

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

be three vectors. A vector

v

in the plane of

a

and

b

, whose projection on

c

\frac{1}{3}

, is given by

Answer: (C)

Since

v

lies in the plane of

a

and

b

, we can write

v = λ a + μ b = λ (\hat{i} + \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k}) = (λ + μ) \hat{i} + (λ - μ) \hat{j} + (λ + μ) \hat{k}

.

The projection of

v

c

\frac{v \cdot c}{∣ c ∣} = \frac{1}{3}

.

Computing:

v \cdot c = (λ + μ) - (λ - μ) - (λ + μ) = μ - λ

.

Since

∣ c ∣ = 3

, we have

\frac{μ - λ}{3} = \frac{1}{3}

, giving

μ - λ = 1

.

Taking

μ = 2

gives

λ = 1

, so

v = 3 \hat{i} - \hat{j} + 3 \hat{k}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option