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MHTCET 22 Apr 2025 Shift 2 Paper

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Question : 7 of 150

Marks: +1, -0

Given

\vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})

\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}

will be

10
12 c
18
20

Solution:

Concept:

The dot product of two vectors is the sum of products of their corresponding components.

Explanation:

First, add

\vec{A}

and

\vec{B}

component-wise.

\vec{A}+\vec{B} = (2+3)\hat{i} + (-3+1)\hat{j} + (1-2)\hat{k}

\vec{A}+\vec{B} = 5\hat{i} - 2\hat{j} - \hat{k}

Write it as

(5,\,-2,\,-1)

.

Given

\vec{C} = 3\hat{i} + 2\hat{j} + \hat{k}

, write it as

(3,\,2,\,1)

.

Now take the dot product:

(\vec{A}+\vec{B})\cdot\vec{C} = (5)(3) + (-2)(2) + (-1)(1)

= 15 - 4 - 1 = 10

Answer:

(\vec{A}+\vec{B})\cdot\vec{C} = 10

Therefore, the correct option is A. 10.

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