Let $\vec{\alpha}=3 \hat{i}+\hat{j}$ and $\vec{\beta}=2 \hat{i}-\hat{j}+3 \hat{k}$. If $\vec{\beta}=\overrightarrow{\beta_{1}}-\overrightarrow{\beta_{2}}$, where $\overrightarrow{\beta_{1}}$ is parallel to $\vec{\alpha}$ and $\overrightarrow{\beta_{2}}$ is perpendicular to $\vec{\alpha}$, then $\overrightarrow{\beta_{1}} \times \overrightarrow{\beta_{2}}$ is equal to:

(A) $\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$

(B) $\frac{1}{2}(3 \hat{i}-9 \hat{j}+5 \hat{k})$

(C) $-3 \hat{i}+9 \hat{j}+5 \hat{k}$

(D) $3 \hat{i}-9 \hat{j}-5 \hat{k}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Vector Algebra

Let

α = 3 \hat{i} + \hat{j}

and

β = 2 \hat{i} - \hat{j} + 3 \hat{k}

. If

β = β_{1} - β_{2}

, where

β_{1}

is parallel to

α

and

β_{2}

is perpendicular to

α

, then

β_{1} \times β_{2}

is equal to:

Answer: (A)

Since

β_{1}

is parallel to

α

, write

β_{1} = λ (3 \hat{i} + \hat{j})

. Then

β_{2} = β_{1} - β = (3 λ - 2) \hat{i} + (λ + 1) \hat{j} - 3 \hat{k}

. Requiring

β_{2} ⊥ α

gives

3 (3 λ - 2) + (λ + 1) = 10 λ - 5 = 0

, so

λ = \frac{1}{2}

. Hence

β_{1} = \frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}

and

β_{2} = - \frac{1}{2} \hat{i} + \frac{3}{2} \hat{j} - 3 \hat{k}

, and their cross product is

β_{1} \times β_{2} = - \frac{3}{2} \hat{i} + \frac{9}{2} \hat{j} + \frac{5}{2} \hat{k} = \frac{1}{2} (- 3 \hat{i} + 9 \hat{j} + 5 \hat{k})

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option