Advertisements
Advertisements
प्रश्न
Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\] is perpendicular to the normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .\]
Advertisements
उत्तर
\[\text{ Direction ratios of the given line } \frac{x - 1}{2}=\frac{y - 1}{3}=\frac{z - 1}{k} \text{ are proportional to 2, 3,k.} \]
\[\text{ Direction ratios of the normal to the plane } \vec{r} .\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)= 4 \text{ are } 2, 3, 4.\]
\[\text{ Given that these two are perpendicular } .\]
\[ \Rightarrow \left( 2 \right) \left( 2 \right) + \left( 3 \right) \left( 3 \right) + \left( k \right)\left( 4 \right) = 0........................... (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 4 + 9 + 4k = 0\]
\[ \Rightarrow 13 + 4k = 0\]
\[ \Rightarrow k = \frac{- 13}{4}\]
