if | {a} | =7, | {b} | =11 and | {a}+{b} | =10{3}, then | {a}-{b} | is equal to

Correct Answer is : 2102{10}210

Correct Answer is : 2102{10}210

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UPSC NDA I 2015 Mathematics Paper

Question : 102 of 120

Marks: +1, -0

| \vec{a} | =7, | \vec{b} | =11

| \vec{a}+\vec{b} | =10\sqrt{3}

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

is equal to

Solution:

Concept:

Use the vector identities for magnitude of sum and difference:

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

and

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

Explanation:

Given:

|\vec{a}|=7

|\vec{b}|=11

|\vec{a}+\vec{b}|=10\sqrt{3}

Step 1: Square the given sum magnitude:

|\vec{a}+\vec{b}|^2 = (10\sqrt{3})^2 = 300

Also,

|\vec{a}|^2=49

|\vec{b}|^2=121

Step 2: Find the dot product using the sum identity:

|\vec{a}+\vec{b}|^2 = 49 + 121 + 2\vec{a}\cdot\vec{b}

So,

300 = 170 + 2\vec{a}\cdot\vec{b}

Thus,

2\vec{a}\cdot\vec{b} = 130

⇒

\vec{a}\cdot\vec{b}=65

Step 3: Use the difference identity:

|\vec{a}-\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} = 49+121-130 = 40

Therefore,

|\vec{a}-\vec{b}| = \sqrt{40} = 2\sqrt{10}

Answer:

2\sqrt{10}

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