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MHTCET 22 Apr 2025 Shift 1 Paper

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Question : 103 of 150

Marks: +1, -0

If

\bar{a}, \bar{b}, \bar{c}

are three unit vectors such that

|\overline{a}+\overline{b}|^{2}+|\overline{a}+\overline{c}|^{2}=8

|\overline{a}+3 \overline{b}|^{2}+|\overline{a}+3 \overline{c}|^{2}=

26
32
22
36

Solution:

Concept:

Use the expansion

|\bar{x}+\bar{y}|^{2}=|\bar{x}|^{2}+|\bar{y}|^{2}+2\bar{x}\cdot\bar{y}

and the given condition to find

\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}

.

Explanation:

Since

\bar{a},\bar{b},\bar{c}

are unit vectors,

|\bar{a}|^{2}=|\bar{b}|^{2}=|\bar{c}|^{2}=1

.

Given:

|\bar{a}+\bar{b}|^{2}+|\bar{a}+\bar{c}|^{2}=8

.

Expanding:

(1+1+2\bar{a}\cdot\bar{b})+(1+1+2\bar{a}\cdot\bar{c})=8

4+2(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c})=8

2(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c})=4

\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}=2

Now find

|\bar{a}+3\bar{b}|^{2}+|\bar{a}+3\bar{c}|^{2}

.

Expanding:

(1+9+6\bar{a}\cdot\bar{b})+(1+9+6\bar{a}\cdot\bar{c})

=20+6(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c})

=20+6(2)=32

Answer:

|\bar{a}+3\bar{b}|^{2}+|\bar{a}+3\bar{c}|^{2}=32

Correct option: B. 32

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