If {a}=2 {i}+2 {j}+3 {k}, {b}=-{i}+2 {j}+{k} and {c}=3 {i}+{j} such that {b}+λ {a} is perpendicular to {c}, then λ is

1/8 According to the given condition, we get {aligned} & ({b}+λ {a}) · {c}=0 \\ & [(-1+2 λ) {{i}}+(2+2 λ) {{j}}+(1+3 λ) {{k}}] ·(3 {{i}}+{{j}})=0 \\ & 3(-1+2 λ)+(2+2 λ)=0 \\ & -3+6 λ+2+2 λ=0 \\ & λ=1/8 {aligned}

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MHTCET 10 May 2024 Shift 2 Paper

Question : 141 of 150

Marks: +1, -0

\overline{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}

and

\bar{c}=3 \hat{i}+\hat{j}

such that

\overline{b}+\lambda \overline{a}

is perpendicular to

\overline{c}

, then

\lambda

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{6}$
$\frac{1}{8}$
According to the given condition, we get
$\begin{aligned} & (\overline{b}+\lambda \overline{a}) \cdot \overrightarrow{c}=0 \\ & \Rightarrow[(-1+2 \lambda) \hat{\mathrm{i}}+(2+2 \lambda) \hat{\mathrm{j}}+(1+3 \lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=0 \\ & \Rightarrow 3(-1+2 \lambda)+(2+2 \lambda)=0 \\ & \Rightarrow-3+6 \lambda+2+2 \lambda=0 \\ & \Rightarrow \lambda=\frac{1}{8} \end{aligned}$

Validate

Solution:

Concept:

Two vectors are perpendicular if their dot product is zero.

Explanation:

Let

\overline{x} = \overline{b} + \lambda \overline{a}

Then

\overline{x} = (-\hat{i}+2\hat{j}+\hat{k}) + \lambda(2\hat{i}+2\hat{j}+3\hat{k})

So,

\overline{x} = (-1+2\lambda)\hat{i} + (2+2\lambda)\hat{j} + (1+3\lambda)\hat{k}

Since

\overline{x}

is perpendicular to

\overline{c} = 3\hat{i}+\hat{j}

, we have

\overline{x} \cdot \overline{c} = 0

(-1+2\lambda)(3) + (2+2\lambda)(1) + (1+3\lambda)(0) = 0

-3 + 6\lambda + 2 + 2\lambda = 0

8\lambda - 1 = 0

\lambda = \frac{1}{8}

Answer:

\lambda = \frac{1}{8}

, which is option D.

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