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Mathematics Question on Angle between two lines

The angle between the lines r⃗=3i^−2j^+1k^+μ(4i^+6j^+12k^)\vec{r} = 3\hat{i} - 2\hat{j} + 1\hat{k} + \mu (4\hat{i} + 6\hat{j} + 12\hat{k}) and r⃗=7i^−3j^+9k^+λ(5i^+8j^−4k^)\vec{r} = 7\hat{i} - 3\hat{j} + 9\hat{k} + \lambda (5\hat{i} + 8\hat{j} - 4\hat{k}) is:

A

cos⁡−1107105\cos^{-1} \frac{10}{\sqrt{7}\sqrt{105}}

B

cos⁡−1572\cos^{-1} \frac{5}{72}

C

cos⁡−1235\cos^{-1} \frac{2}{35}

D

cos⁡−1798\cos^{-1} \frac{7}{98}

Answer

cos⁡−1107105\cos^{-1} \frac{10}{\sqrt{7}\sqrt{105}}

Explanation

Solution

The angle between two lines is determined by the angle between their direction vectors. For the given lines, the direction vectors are:

d1⃗=4i^+6j^+12k^\vec{d_1} = 4\hat{i} + 6\hat{j} + 12\hat{k}, d2⃗=5i^+8j^−4k^\vec{d_2} = 5\hat{i} + 8\hat{j} - 4\hat{k}.

The formula for the cosine of the angle between two vectors is:

cos⁡θ=d1⃗⋅d2⃗∥d1⃗∥∥d2⃗∥.\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{\|\vec{d_1}\| \|\vec{d_2}\|}.

Compute the dot product d1⃗⋅d2⃗\vec{d_1} \cdot \vec{d_2}:

d1⃗⋅d2⃗=(4)(5)+(6)(8)+(12)(−4).\vec{d_1} \cdot \vec{d_2} = (4)(5) + (6)(8) + (12)(-4).

d1⃗⋅d2⃗=20+48−48=20.\vec{d_1} \cdot \vec{d_2} = 20 + 48 - 48 = 20.

Compute the magnitudes of d1⃗\vec{d_1} and d2⃗\vec{d_2}. The magnitude of d1⃗\vec{d_1} is:

∥d1⃗∥=(4)2+(6)2+(12)2=16+36+144=196=14.\|\vec{d_1}\| = \sqrt{(4)^2 + (6)^2 + (12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14.

The magnitude of d2⃗\vec{d_2} is:

∥d2⃗∥=(5)2+(8)2+(−4)2=25+64+16=105.\|\vec{d_2}\| = \sqrt{(5)^2 + (8)^2 + (-4)^2} = \sqrt{25 + 64 + 16} = \sqrt{105}.

Substitute into the cosine formula:

cos⁡θ=d1⃗×d2⃗∥d1⃗∥∥d2⃗∥.\cos \theta = \frac{\vec{d_1} \times \vec{d_2}}{\|\vec{d_1}\| \|\vec{d_2}\|}.

cos⁡θ=2014×105=107×105.\cos \theta = \frac{20}{14 \times \sqrt{105}} = \frac{10}{\sqrt{7} \times \sqrt{105}}.

Thus, the angle between the lines is:

θ=cos⁡−1(107105).\theta = \cos^{-1} \left( \frac{10}{\sqrt{7} \sqrt{105}} \right).