Solve this following

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

$\left(\frac{1}{2}, 4,-2\right)$
$\left(-\frac{1}{2}, 4,0\right)$
$(1,3,1)$
$(1,5,1)$

Correct Option: , 2

Solution:

$4 \hat{\mathrm{i}}+\left(3-\lambda_{2}\right) \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=4 \hat{\mathrm{i}}+2 \lambda_{1} \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$

$\Rightarrow 3-\lambda_{2}=2 \lambda_{1} \Rightarrow 2 \lambda_{1}+\lambda_{2}=3$ ..........(1)

Given $\vec{a} \cdot \vec{c}=0$

$\Rightarrow 6+6 \lambda_{1}+3\left(\lambda_{3}-1\right)=0$

$\Rightarrow 2 \lambda_{1}+\lambda_{3}=-1$ ....................(2)

Now $\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\left(\lambda_{1}, 3-2 \lambda_{1},-1-2 \lambda_{1}\right)$

Now check the options, option (2) is correct

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