CUET Mathematics 2024 - The angle between two lines whose direction ratios are propotional to 1,1,-2 and (sqrt(3)-1),(-sqrt(3)-1),-4 is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Vector Algebra

Medium

The angle between two lines whose direction ratios are propotional to $1,1,-2$ and $(\sqrt{3}-1),(-\sqrt{3}-1),-4$ is :

\frac{\pi}{3}

\pi

\frac{\pi}{6}

\frac{\pi}{2}

✅ Correct Option: 1

We need to find the angle between two lines with direction ratios proportional to

(1, 1, -2)

and

(\sqrt{3}-1, -\sqrt{3}-1, -4)

For any line with direction ratios

(a, b, c)

, the unit vector is obtained by dividing by the magnitude

\sqrt{a^2+b^2+c^2}

For the first line

(1, 1, -2)

Magnitude =

\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}

Unit vector

\vec{a} = \dfrac{1}{\sqrt{6}}(1, 1, -2)

For the second line

(\sqrt{3}-1, -\sqrt{3}-1, -4)

Magnitude =

\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + (-4)^2}

Simplifying:

(\sqrt{3}-1)^2 = 4 - 2\sqrt{3}

(-\sqrt{3}-1)^2 = 4 + 2\sqrt{3}

(-4)^2 = 16

Magnitude =

\sqrt{(4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}

Unit vector

\vec{b} = \dfrac{1}{2\sqrt{6}}(\sqrt{3}-1, -\sqrt{3}-1, -4)

To find the angle, we calculate the dot product of the unit vectors:

\cos(\theta) = \vec{a} \cdot \vec{b}

\cos(\theta) = \dfrac{1}{\sqrt{6}} \cdot \dfrac{\sqrt{3}-1}{2\sqrt{6}} + \dfrac{1}{\sqrt{6}} \cdot \dfrac{-\sqrt{3}-1}{2\sqrt{6}} + \dfrac{-2}{\sqrt{6}} \cdot \dfrac{-4}{2\sqrt{6}}

\cos(\theta) = \dfrac{1}{12}[(\sqrt{3}-1) + (-\sqrt{3}-1) + 8]

\cos(\theta) = \dfrac{1}{12}[-2 + 8] = \dfrac{6}{12} = \dfrac{1}{2}

Since

\cos(\theta) = \dfrac{1}{2}

, we have

\theta = \cos^{-1}(\dfrac{1}{2}) = 60°

Therefore, the angle between the two lines is

60° = \dfrac{\pi}{3}