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MHTCET 11 May 2024 Shift 2 Paper

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Question : 147 of 150

Marks: +1, -0

If the vectors

\overline{a}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{b}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}

\overline{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu \hat{\mathrm{k}}

are mutually orthogonal, then

(\lambda, \mu) \equiv

(-3,2)
$(2,-3)$
(-2,3)
(3, -2)

Solution:

Concept:

For mutually orthogonal vectors, the dot product of every pair is zero.

Explanation:

Given

\bar{a} \cdot \bar{c} = 0

:

(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})\cdot(\lambda\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu\hat{\mathrm{k}}) = 0

So,

\lambda - 1 + 2\mu = 0

, giving

\lambda + 2\mu = 1

.

Also,

\bar{b} \cdot \bar{c} = 0

:

(2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+\hat{\mathrm{k}})\cdot(\lambda\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu\hat{\mathrm{k}}) = 0

So,

2\lambda + 4 + \mu = 0

, giving

2\lambda + \mu = -4

.

Solve the two equations:

From

\lambda + 2\mu = 1

, write

\lambda = 1 - 2\mu

.

Substitute into

2\lambda + \mu = -4

:

2(1 - 2\mu) + \mu = -4

2 - 4\mu + \mu = -4

-3\mu = -6

, hence

\mu = 2

.

Then

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

.

Therefore,

(\lambda, \mu) = (-3, 2)

.

Answer:

(\lambda, \mu) = (-3, 2)

, which matches option A.

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