The angle between two lines whose direction ratio are proportional to (-, -2, ) and (-2, ), where R \0\, is Option 1: ^(-1)(1/3) Option 2: ^(-1)(-1/6) Option 3: ()/2 Option 4: ()/4

Answer: ^(-1)(-1/6). Solution: The dot product is 2^2-2^2+^2=^2, while each direction vector has magnitude √6 . Hence the cosine of the angle is (^2)/(6^2)=1/6. Note: The computation yields ^(-1)(1/6); only the option carrying magnitude 1/6 matches, so we have taken it as intended.

Geometry

3D Geometry

Medium

core

The angle between two lines whose direction ratio are proportional to $(-\theta, -2\theta, \theta)$ and $(-2\theta, \theta, \theta)$ , where $\theta \in \mathbb{R} \sim \{0\}$ , is

$\cos^{-1}\left(\frac{1}{3}\right)$

$\cos^{-1}\left(-\frac{1}{6}\right)$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

✅ Correct Option: 2

The dot product is $2\theta^2-2\theta^2+\theta^2=\theta^2$ , while each direction vector has magnitude $\sqrt{6}\,|\theta|$ . Hence the cosine of the angle is $\frac{\theta^2}{6\theta^2}=\frac{1}{6}$ .

Note: The computation yields $\cos^{-1}(1/6)$ ; only the option carrying magnitude $1/6$ matches, so we have taken it as intended.

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The angle between the two planes $x + y - z = 3$ and $3x + 2y + z = 5$ is :

The angle between two lines whose direction ratio are proportional to (−θ,−2θ,θ)(-\theta, -2\theta, \theta)(−θ,−2θ,θ) and (−2θ,θ,θ)(-2\theta, \theta, \theta)(−2θ,θ,θ), where θ∈R∼{0}\theta \in \mathbb{R} \sim \{0\}θ∈R∼{0}, is

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CUET Mathematics Past Year Questions (Topic-Wise)

The angle between two lines whose direction ratio are proportional to $(-\theta, -2\theta, \theta)$ and $(-2\theta, \theta, \theta)$ , where $\theta \in \mathbb{R} \sim \{0\}$ , is