The angle between the pair of lines {r} = -3{i} + {j} + 3{k} + λ(3{i} + 5{j} + 4{k}) and {r} = -{i} + 4{j} + 5{k} + ({i} + {j} + 2{k}) is . . . . . . .

Correct Answer is : cos⁡−1(8315)^{-1}({8{3}}{15})cos−1(1583)

Correct Answer is : cos⁡−1(8315)^{-1}({8{3}}{15})cos−1(1583)

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GUJCET 2025 Math Paper

Question : 4 of 40

Marks: +1, -0

\vec{r} = -3\hat{i} + \hat{j} + 3\hat{k} + \lambda(3\hat{i} + 5\hat{j} + 4\hat{k})

\vec{r} = -\hat{i} + 4\hat{j} + 5\hat{k} + \mu(\hat{i} + \hat{j} + 2\hat{k})

is . . . . . . .

Solution:

Concept:

The angle between two lines is determined from their direction vectors using the dot product formula.

Explanation:

The direction vectors are

\vec{b}_1 = 3\hat{i} + 5\hat{j} + 4\hat{k}

and

\vec{b}_2 = \hat{i} + \hat{j} + 2\hat{k}

Use

\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}

Dot product:

\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (5)(1) + (4)(2) = 3 + 5 + 8 = 16

Magnitudes:

|\vec{b}_1| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{50} = 5\sqrt{2}

|\vec{b}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}

Substitute:

\cos \theta = \frac{16}{5\sqrt{2}\sqrt{6}} = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}

Rationalize:

\frac{8}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{15}

Therefore,

\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)

Answer:

\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)

, so the correct option is C.

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