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Vector Algebra question

2021 · 25 Jul · Shift 2 · Q34

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Vector Algebra
2021 · 25 Jul · Shift 2 · Q34

Vector Algebra question

2021 · 25 Jul · Shift 2 · Q34

JEE MainMathematicsVector AlgebraMCQ+4 / −1

If

\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5

and

\left| {\overrightarrow a \times \overrightarrow b } \right|

= 8, then

\left| {\overrightarrow a .\,\overrightarrow b } \right|

is equal to :

A
6
B
4
C
3
D
5

View written solutionFree

Correct answer: A

We use the identity relating dot product and cross product:

$|\vec a \times \vec b|^2 + (\vec a \cdot \vec b)^2 = |\vec a|^2 |\vec b|^2$

Substitute the given values:

$|\vec a| = 2, \quad |\vec b| = 5, \quad |\vec a \times \vec b| = 8$

So,

$8^2 + (\vec a \cdot \vec b)^2 = 2^2 \cdot 5^2$

$64 + (\vec a \cdot \vec b)^2 = 4 \cdot 25$

$64 + (\vec a \cdot \vec b)^2 = 100$

Solve for $(\vec a \cdot \vec b)^2$ :

$(\vec a \cdot \vec b)^2 = 100 - 64 = 36$

$|\vec a \cdot \vec b| = \sqrt{36} = 6$

Hence the correct option is:

$\boxed{\text{A: }6}$

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