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MHTCET 26 Apr 2025 Shift 1 Paper

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Question : 11 of 150

Marks: +1, -0

If

\quad|\vec{a}|=\sqrt{26},|\vec{b}|=7

|\vec{a} \times \vec{b}|=35 \text {, find } \quad \vec{a} \cdot \vec{b}

4
5
6
7

Solution:

Concept:

Use the vector cross and dot product formulas to find the dot product from the given magnitudes and cross product magnitude.

Explanation:

Given

|\vec{a}|=\sqrt{26}

,

|\vec{b}|=7

, and

|\vec{a} \times \vec{b}|=35

.

The magnitude of the cross product is:

|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta

So,

35 = \sqrt{26} \cdot 7 \cdot \sin\theta

\sin\theta = \frac{35}{7\sqrt{26}} = \frac{5}{\sqrt{26}}

Now, find

\cos\theta

using the identity

\sin^2\theta + \cos^2\theta = 1

:

\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{25}{26}} = \frac{1}{\sqrt{26}}

The dot product is:

\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta

\vec{a} \cdot \vec{b} = \sqrt{26} \cdot 7 \cdot \frac{1}{\sqrt{26}} = 7

Answer:

\vec{a} \cdot \vec{b} = 7

Correct option: D. 7

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