Q100 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

The direction cosines of the normal to the plane containing the lines having direction ratios $1,2,1$ and $4,5,-3$ are

  • Correct answer
    A
    $\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}$
  • B
    $\frac{1}{\sqrt{2}}, 0, \frac{-1}{\sqrt{2}}$
  • C
    $\frac{5}{\sqrt{41}}, \frac{-4}{\sqrt{41}}, 0$
  • D
    $\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}, 0$

Solution

Let $\mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}_2=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ Normal vector to plane $\mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & 2 & 1 \\ 4 & 5 & -3\end{array}\right|$ $=\hat{\mathbf{i}}(-6-5)-\hat{\mathbf{j}}(-3-4)+\hat{\mathbf{k}}(5-8)$ $=-11 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ $\therefore$ Direction cosines of normal to the plane $=\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}},$ $\frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}}=\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}$

Hence, the correct answer is option: 'A'.

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