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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 139 of 150

Marks: +1, -0

\bar{a}=4 \hat{i}+\hat{j}+\hat{k}^{\prime} \bar{b}=2 \hat{i}+\hat{j}+2 \hat{k}

\bar{c}=3 \hat{i}+4 \hat{j}+5 \hat{k}

(\bar{a}+\bar{b}) \cdot(\bar{b}+\bar{c})=

Solution:

Concept:

Use vector addition to combine components, then apply the dot product formula:

(\bar{x} \cdot \bar{y}) = x_1 y_1 + x_2 y_2 + x_3 y_3

Explanation:

Add the corresponding components of

\bar{a}

and

\bar{b}

\bar{a}+\bar{b} = (4 \hat{i}+\hat{j}+\hat{k}) + (2 \hat{i}+\hat{j}+2 \hat{k})

\bar{a}+\bar{b} = 6 \hat{i}+2 \hat{j}+3 \hat{k}

Next, add

\bar{b}

and

\bar{c}

\bar{b}+\bar{c} = (2 \hat{i}+\hat{j}+2 \hat{k}) + (3 \hat{i}+4 \hat{j}+5 \hat{k})

\bar{b}+\bar{c} = 5 \hat{i}+5 \hat{j}+7 \hat{k}

Now take the dot product:

(\bar{a}+\bar{b}) \cdot (\bar{b}+\bar{c}) = (6 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot (5 \hat{i}+5 \hat{j}+7 \hat{k})

Multiply corresponding components and add:

= 6(5) + 2(5) + 3(7)

= 30 + 10 + 21

= 61

Answer:

(\bar{a}+\bar{b}) \cdot (\bar{b}+\bar{c}) = 61

Correct option: C

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