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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2024 · Paper 1 · Mathematics · Three Dimensional Geometry

Let the foot of perpendicular from a point P(1,2,-1) to the...

Let the foot of perpendicular from a point P(1,2,−1)\mathrm{P}(1,2,-1) to the straight line L:x1=y0=z−1\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{0}=\frac{\mathrm{z}}{-1} be NN. Let a line be drawn from PP parallel to the plane x+y+2z=0x+y+2 z=0 which meets LL at point QQ. If α\alpha is the acute angle between the lines PN and PQ , then cos⁡α\cos \alpha is equal to \qquadLet the foot of perpendicular from a point P(1,2,-1) to the straight line L: x 1 = y 0 = z -1 be N . Let a line beLet the foot of perpendicular from a point P(1,2,-1) to the straight line L: x 1 = y 0 = z -1 be N . Let a line be

Options

  1. A15\frac{1}{\sqrt{5}}
  2. B32\frac{\sqrt{3}}{2}
  3. C13\frac{1}{\sqrt{3}}
  4. D123\frac{1}{2 \sqrt{3}}

Correct answer

C. 13\frac{1}{\sqrt{3}}

Step-by-step solution

Let x1=y0=z−1=λ⇒N(λ,0,−λ)b→=i^−k^\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}=\lambda\Rightarrow \mathrm{N}(\lambda, 0,-\lambda)\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{k}} Let the foot of perpendicular from a point P(1,2,-1) to the straight line L: x 1 = y 0 = z -1 be N . Let a line be ∵PN→⋅b→=0⇒1(1−λ)−(λ−1)=0⇒λ=1⇒N(1,0,−1)Let⁡Q(μ,0,−μ)∵n→=i^+j^+2k^\because \overrightarrow{\mathrm{PN}} \cdot \overrightarrow{\mathrm{b}}=0\Rightarrow 1(1-\lambda)-(\lambda-1)=0 \Rightarrow \lambda=1\Rightarrow \mathrm{N}(1,0,-1)\operatorname{Let} \mathrm{Q}(\mu, 0,-\mu)\because \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} Now, Let the foot of perpendicular from a point P(1,2,-1) to the straight line L: x 1 = y 0 = z -1 be N . Let a line be ∵PQ→⋅n→=0⇒μ=−1⇒Q(−1,0,1)PN→=2j^\because \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{n}}=0\Rightarrow \mu=-1\Rightarrow \mathrm{Q}(-1,0,1)\overrightarrow{\mathrm{PN}}=2 \hat{\mathrm{j}} and PQ→=2i^+2j^−2k^⇒cos⁡α=13\overrightarrow{\mathrm{PQ}}=2 \hat{i}+2 \hat{j}-2 \hat{k}\Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}
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