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Vector Algebra question

2012 · Shift 0 · Q28

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JEE Main
Mathematics
Vector Algebra
2012 · Shift 0 · Q28

Vector Algebra question

2012 · Shift 0 · Q28

JEE MainMathematicsVector AlgebraMCQ+4 / −1

Let

\overrightarrow a

and

\overrightarrow b

be two unit vectors. If the vectors

\,\overrightarrow c = \widehat a + 2\widehat b

and

\overrightarrow d = 5\widehat a - 4\widehat b

are perpendicular to each other, then the angle between

\overrightarrow a

and

\overrightarrow b

is :

A
${\pi \over 6}$
B
${\pi \over 2}$
C
${\pi \over 3}$
D
${\pi \over 4}$

View written solutionFree

Correct answer: C

Let the angle between the unit vectors $\vec a$ and $\vec b$ be $\theta$ .
Given:

\vec c = \vec a + 2\vec b, \qquad \vec d = 5\vec a - 4\vec b

Since $\vec c$ and $\vec d$ are perpendicular,

\vec c \cdot \vec d = 0

Compute the dot product:

(\vec a + 2\vec b) \cdot (5\vec a - 4\vec b) = 0

Expanding,

5\vec a\cdot \vec a - 4\vec a\cdot \vec b + 10\vec b\cdot \vec a - 8\vec b\cdot \vec b = 0

Since dot product is commutative,

\vec a\cdot \vec b = \vec b\cdot \vec a

and because $\vec a, \vec b$ are unit vectors,

\vec a\cdot \vec a = 1, \qquad \vec b\cdot \vec b = 1

So,

5 - 4(\vec a\cdot \vec b) + 10(\vec a\cdot \vec b) - 8 = 0

5 - 8 + 6(\vec a\cdot \vec b) = 0

-3 + 6(\vec a\cdot \vec b) = 0

\vec a\cdot \vec b = \frac{1}{2}

Now,

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

because both are unit vectors. Thus,

\cos\theta = \frac{1}{2}

which gives

\theta = \frac{\pi}{3}

Checking options:

A: $\frac{\pi}{6}$ ❌
B: $\frac{\pi}{2}$ ❌
C: $\frac{\pi}{3}$ ✅
D: $\frac{\pi}{4}$ ❌

Therefore, the correct answer is $\boxed{\frac{\pi}{3}}$ .

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