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प्रश्न
Find the distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k} \right)\] and the plane \[\vec{r} . \left( \hat{i} - \hat{j} + \hat{k} \right) = 5 .\]
उत्तर
\[\text{ The given equation of the line is} \]
\[ \vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda \left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right)\]
\[ \Rightarrow \vec{r} = \left( 2 + 3\lambda \right) \hat{i} + \left( - 1 + 4\lambda \right) \hat{j} + \left( 2 + 2\lambda \right) \hat{k} \]
\[\text{ The coordinates of any point on this line are of the form} \left( 2 + 3\lambda \right) \hat{i} + \left( - 1 + 4\lambda \right) \hat{j} + \left( 2 + 2\lambda \right) \hat{k} . or\left( 2 + 3\lambda, - 1 + 4\lambda, 2 + 2\lambda \right)\]
\[\text{ Since this point lies on the plane } \vec{r} .\left( \hat{i} - \hat{j} + \hat{k} \right)= 5,\]
\[\left[ \left( 2 + 3\lambda \right) \hat{i} + \left( - 1 + 4\lambda \right) \hat{j} + \left( 2 + 2\lambda \right) \hat{k} \right] . \left( \hat{i} - \hat{j} + \hat{k} \right) = 5\]
\[ \Rightarrow 2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda - 5 = 0\]
\[ \Rightarrow \lambda = 0\]
\[\text{ So, the coordinates of the point are } \]
\[\left( 2 + 3\lambda, - 1 + 4\lambda, 2 + 2\lambda \right)\]
\[ = \left( 2 + 0, - 1 + 0, 2 + 0 \right)\]
\[ = \left( 2, - 1, 2 \right)\]
\[\text{ Distance between (2, -1, 2) and (-1, -5, -10)} \]
\[ = \sqrt{\left( - 1 - 2 \right)^2 + \left( - 5 + 1 \right)^2 + \left( - 10 - 2 \right)^2}\]
\[ = \sqrt{9 + 16 + 144}\]
\[ = 13 \text{ units } \]
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