Visual SolutionEasy

Question

The angle between the vectors

\vec{a} = \hat{i} + \hat{j}

and

\vec{b} = \hat{j} + \hat{k}

is:

(A)

\dfrac{\pi}{3}

(B)

\dfrac{\pi}{4}

(C)

\dfrac{\pi}{6}

(D)

\dfrac{\pi}{2}

Solution Path

Dot product = 1, magnitudes = sqrt(3) each, cos theta = 1/3.

01Question Setup

1/4

Find the angle between

\vec{a} = \hat{i} + \hat{j} + \hat{k}

and

\vec{b} = \hat{i} - \hat{j} + \hat{k}

\theta = ;?

02Dot Product

2/4

\vec{a} \cdot \vec{b} = 1(1) + 1(-1) + 1(1) = 1

|\vec{a}| = \sqrt{3}

|\vec{b}| = \sqrt{3}

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

03Compute AngleKEY INSIGHT

3/4

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{3}

. So

\theta = \cos^{-1}(1/3)

\theta = \cos^{-1}\frac{1}{3}

04Final Answer

4/4

\theta = \cos^{-1}(1/3) \approx 70.5^\circ

\boxed{\cos^{-1}\frac{1}{3}}

Concepts from this question1 concepts unlocked

★

Dot Product and Angle Between Vectors

EASY

The dot product a.b = |a||b|cos theta directly gives the angle between two vectors. Compute the dot product from components, then divide by the product of magnitudes to isolate cos theta.

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

Finding the angle between two vectors is one of the most frequently tested operations in JEE. It also underlies orthogonality checks (cos theta = 0) and projection problems.

Angle between vectorsOrthogonality testProjection problemsWork done in physics

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