MHT CET · Maths · Three Dimensional Geometry
The direction cosines of a line which is perpendicular to lines whose direction ratit \(3,-2,4\) and \(1,3,-2\) are
- A \(\frac{-8}{\sqrt{285}}, \frac{-10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
- B \(\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
- C \(\frac{8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
- D \(\frac{4}{\sqrt{297}}, \frac{5}{\sqrt{297}}, \frac{16}{\sqrt{297}}\)
Answer & Solution
Correct Answer
(B) \(\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
Step-by-step Solution
Detailed explanation
111.(B)
Let \(\bar{a}=3 \hat{i}-2 \hat{j}+4 \hat{k}\) and \(\bar{b}=1 \hat{i}+3 \hat{j}-2 \hat{k}\)
\(\begin{aligned}
\bar{a} \times \bar{b} &=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -2 & 4 \\
1 & 3 & -2
\end{array}\right| \\
\bar{a} \times \bar{b} &=(4-12) \hat{i}-(-6-4) \hat{j}+(9+2) \hat{k} \\
&=-8 \hat{i}+10 \hat{j}+11 \hat{k} \\
|\bar{a} \times \bar{b}| &=\sqrt{(-8)^{2}+(10)^{2}+(11)^{2}}=\sqrt{285}
\end{aligned}\)
Direction cosines are : \(\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
Let \(\bar{a}=3 \hat{i}-2 \hat{j}+4 \hat{k}\) and \(\bar{b}=1 \hat{i}+3 \hat{j}-2 \hat{k}\)
\(\begin{aligned}
\bar{a} \times \bar{b} &=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -2 & 4 \\
1 & 3 & -2
\end{array}\right| \\
\bar{a} \times \bar{b} &=(4-12) \hat{i}-(-6-4) \hat{j}+(9+2) \hat{k} \\
&=-8 \hat{i}+10 \hat{j}+11 \hat{k} \\
|\bar{a} \times \bar{b}| &=\sqrt{(-8)^{2}+(10)^{2}+(11)^{2}}=\sqrt{285}
\end{aligned}\)
Direction cosines are : \(\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}\)
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