Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. A vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$, is

(A) $4\hat{i}-\hat{j}+4\hat{k}$

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

(C) $2\hat{i}+\hat{j}-2\hat{k}$

(D) $4\hat{i}+\hat{j}-4\hat{k}$

JEE Advanced · 2006 · 09 Apr 2006 · Vector Algebra

Let

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

and

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

. A vector in the plane of

a

and

b

whose projection on

c

\frac{1}{3}

, is

Answer: (A)

A vector in the plane of

a

and

b

can be written as

r = a + λ b

r = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) = (1 + λ) \hat{i} + (2 - λ) \hat{j} + (1 + λ) \hat{k}

The projection of

r

c

is:

\frac{r \cdot c}{∣ c ∣} = \frac{( 1 + λ ) - ( 2 - λ ) - ( 1 + λ )}{3} = \frac{λ - 2}{3}

Given this equals

\frac{1}{3}

\frac{λ - 2}{3} = \frac{1}{3} \Rightarrow λ = 3

Therefore:

r = (1 + 3) \hat{i} + (2 - 3) \hat{j} + (1 + 3) \hat{k} = 4 \hat{i} - \hat{j} + 4 \hat{k}

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option