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If the distance between the plane, 23x-10y-2z+48=0 and the plane containing the lines \frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3} \: \: and \: \: \frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }\left ( \lambda \epsilon \textbf{R} \right ) is equal to \frac{k}{\sqrt{633}}, then k is equal to ________.

Option: 1

14


Option: 2

10


Option: 3

13


Option: 4

3


Answers (1)

best_answer

 

 

Distance of a Point From a Plane -
Cartesian Form

Let P(x1, y1, z1) be the given point with position vector \vec{\mathbf a} and ax + by + cz + d = 0 be the Cartesian equation of the given plane. Then

\\ \mathrm{\;\;}\vec{ a} =x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k} \\ \overrightarrow =\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}

Hence, from Vector form of the perpendicular from P to the plane is

\\\left|\frac{\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|  

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Lines must be intersecting

\\\Rightarrow(2 a-1,4 a+3,3 a-1)=(2 b-3,6 b-2, \lambda b+1)\\\Rightarrow 2 a-1=2 b-3,4 a+3=6 b-2,3 a-1=\lambda b+1\\\Rightarrow \mathrm{b}=\frac{1}{2}, \mathrm{a}=-\frac{1}{2}, \lambda=-7\\\text {distance of plane contains given lines from given plane is same }\\\text{ as distance between point }(-3,-2,1)\;\text{from given plane. }\\ \text{Required distance equal to}\\\frac{|-69+20-2+48|}{\sqrt{529+100+4}}=\frac{3}{\sqrt{633}}=\frac{\mathrm{k}}{\sqrt{633}} \Rightarrow \mathrm{k}=3

 

Posted by

Ritika Kankaria

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