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प्रश्न
The direction cosines of the line x - y + 2z = 5 and 3x + y + z = 6 are
विकल्प
\[\frac{-3}{5\sqrt{2}},\frac{5}{5\sqrt{2}},\frac{4}{5\sqrt{2}}\]
\[\frac{3}{5\sqrt{2}},\frac{-5}{5\sqrt{2}},\frac{4}{5\sqrt{2}}\]
\[\frac{3}{5\sqrt{2}},\frac{5}{5\sqrt{2}},\frac{4}{5\sqrt{2}}\]
\[\frac{3}{5\sqrt2},\frac{5}{5\sqrt2},\frac{-4}{5\sqrt2}\]
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उत्तर
\[\frac{-3}{5\sqrt{2}},\frac{5}{5\sqrt{2}},\frac{4}{5\sqrt{2}}\]
Explanation:
Given equations of planes are
x − y + 2z = 5
3x + 2y + z = 6
The vectors normal to the given planes are
\[\mathrm{n}_1=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}\] and
\[\mathrm{n_2=3\hat{i}+\hat{j}+\hat{k}}\]
\[\mathrm{n=n_1\times n_2}\]
\[= \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix}\]
\[=-3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}\]
\[\therefore\quad\hat{\mathrm{n}}=\frac{-3}{5\sqrt{2}}\hat{\mathrm{i}}+\frac{5}{5\sqrt{2}}\hat{\mathrm{j}}+\frac{4}{5\sqrt{2}}\hat{\mathrm{k}}\]
