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MHTCET 11 May 2024 Shift 2 Paper

Question : 113 of 150

Marks: +1, -0

\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})

\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})

(2 \hat{a}-\hat{b}) \cdot[(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})]

Solution:

Concept:

\hat{a}

and

\hat{b}

are perpendicular unit vectors, so the vector triple product simplifies neatly.

Explanation:

First, compute

\hat{a} \cdot \hat{b}

\hat{a} \cdot \hat{b} = \frac{1}{7\sqrt{10}}(3 \cdot 2 + 0 \cdot 3 + 1 \cdot (-6)) = 0.

Thus,

\hat{a} \perp \hat{b}

, and both are unit vectors.

Using the vector triple product identity:

(\hat{a} \times \hat{b}) \times (\hat{a} + 2\hat{b}) = \hat{b} - 2\hat{a}.

Substitute into the given expression:

(2\hat{a} - \hat{b}) \cdot (\hat{b} - 2\hat{a}) = -|2\hat{a} - \hat{b}|^2.

Now,

|2\hat{a} - \hat{b}|^2 = 4|\hat{a}|^2 + |\hat{b}|^2 - 4(\hat{a} \cdot \hat{b}) = 4 + 1 - 0 = 5.

Therefore, the value is

-5

Answer:

-5

(Option C)

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