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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 104 of 120

Marks: +1, -0

If

| \vec{a} |

| \vec{b} | =4

| \vec{a}-\vec{b} | =5,

then what is the value of

| \vec{a}+\vec{b} |

?

8
6
5√2
5

Solution:

Concept:

Use the vector identity

|\vec{a}-\vec{b}|^{2}+|\vec{a}+\vec{b}|^{2}=2\big(|\vec{a}|^{2}+|\vec{b}|^{2}\big)

.

Explanation:

Given

|\vec{a}|=3

,

|\vec{b}|=4

, and

|\vec{a}-\vec{b}|=5

.

Substitute these values into the identity:

5^{2}+|\vec{a}+\vec{b}|^{2}=2\times(3^{2}+4^{2})

.

Compute the squares:

25+|\vec{a}+\vec{b}|^{2}=2\times(9+16)=2\times25=50

.

Solve for

|\vec{a}+\vec{b}|^{2}

:

|\vec{a}+\vec{b}|^{2}=50-25=25

.

Take the positive square root:

|\vec{a}+\vec{b}|=\sqrt{25}=5

.

Answer:

5

(Option D).

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