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KCET 2026 Mathematics Solved Paper

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Question : 50 of 60

Marks: +1, -0

If

\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}, \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}

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

\alpha+\beta

is equal to

2
-1
0
1

Solution:

Concept:

The condition

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

implies

\vec{a}\cdot\vec{b}=0

, i.e., the vectors are perpendicular.

Explanation:

Given:

\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}, \quad \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}

Also,

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

.

Square both sides:

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

Using

|\vec{x}|^{2}=\vec{x}\cdot\vec{x}

, we get:

(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=(\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b})

Expanding:

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

Cancelling common terms:

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

4\vec{a}\cdot\vec{b}=0 \implies \vec{a}\cdot\vec{b}=0

Now compute

\vec{a}\cdot\vec{b}

:

\vec{a}\cdot\vec{b}=2\alpha+2\beta-2

Since

\vec{a}\cdot\vec{b}=0

:

2\alpha+2\beta-2=0 \implies \alpha+\beta=1

Answer:

\alpha+\beta=1

, so the correct option is D. 1.

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