Let $\vec{a}=2 \hat{i}+\lambda_{1} \hat{j}+3 \hat{k}, \vec{b}=4 \hat{i}+\left(3-\lambda_{2}\right) \hat{j}+6 \hat{k}$ and $\vec{c}=3 \hat{i}+6 \hat{j}+\left(\lambda_{3}-1\right) \hat{k}$ be three vectors such that $\vec{b}=2 \vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. Then a possible value of $(\lambda_{1}, \lambda_{2}, \lambda_{3})$ is:

(B) $\left(-\frac{1}{2}, 4,0\right)$

(C) $\left(\frac{1}{2}, 4,-2\right)$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Vector Algebra

Let

a = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}, b = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}

and

c = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}

be three vectors such that

b = 2 a

and

a

is perpendicular to

c

. Then a possible value of

(λ_{1}, λ_{2}, λ_{3})

is:

Answer: (B)

From

b = 2 a

, comparing

\hat{j}

components gives

3 - λ_{2} = 2 λ_{1}

, so

λ_{2} = 3 - 2 λ_{1}

. From

a ⊥ c

, the dot product

6 + 6 λ_{1} + 3 (λ_{3} - 1) = 0

simplifies to

λ_{3} = - 1 - 2 λ_{1}

. These leave one free parameter, so the solutions form the family

(λ_{1}, 3 - 2 λ_{1}, - 1 - 2 λ_{1})

. Taking

λ_{1} = - \frac{1}{2}

gives

λ_{2} = 4

and

λ_{3} = 0

, so

(- \frac{1}{2}, 4, 0)

is a valid triple.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option