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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2014 · Mathematics · Three Dimensional Geometry

Find the angle between the line (x+1)/(2)=(y)/(3)=(z-3)/(6) and the...

Find the angle between the line x+12=y3=z−36\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} and the plane 10x+2y−11z=310 \mathrm{x}+2 \mathrm{y}-11 \mathrm{z}=3

Options

  1. Asin⁡−1(821)\sin ^{-1} \left(\frac{8}{21}\right)
  2. Bsin⁡−1(521)\sin ^{-1} \left(\frac{5}{21}\right)
  3. Csin⁡−1(721)\sin ^{-1} \left(\frac{7}{21}\right)
  4. Dsin⁡−1(121)\sin ^{-1} \left(\frac{1}{21}\right)

Correct answer

A. sin⁡−1(821)\sin ^{-1} \left(\frac{8}{21}\right)

Step-by-step solution

Let θ\theta be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have r→=(−i^+3k^)+λ(2i^+3j^+6k^)\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) and r→⋅(\overrightarrow{\mathrm{r}} \cdot(10 \hat{i\mathrm{i}}+2 \hat{j\mathrm{j}}-11 \hat{k\mathrm{k}})=3 Here, b→=2i^+3j^+6k^\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} and n→=10i^+2j^−11k^\overrightarrow{\mathrm{n}}=10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-11 \hat{\mathrm{k}} sin⁡ϕ=∣(2i^+3j^+6k^)⋅(10i^+2j^−11k^)22+32+62102+22+112∣=∣−407×15∣=∣−821∣=821 or ϕ=sin⁡−1(821)\begin{array}{l} \sin \phi=\left|\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k}) \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+11^{2}}}\right|\\ =\left|\frac{-40}{7 \times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21} \text{ or } \phi=\sin ^{-1}\left(\frac{8}{21}\right) \end{array}
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