Question

hello brainliacs

help me in solving this

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Answer 1

Answer :-

$\boxed{ \sf{ \frac{x - 1}{3} = \frac{y - 3}{5} = \frac{z -5 }{ - 1}} } \:$

To Find :-

→ Equation of line .

Step - by - step explanation :-

Let , m be the required equation of line .

$\sf{given \: line} \: \\ \sf{let } \\ \small\sf{l = \frac{x - 2}{1} = \frac{y - 2}{ - 1} =\frac{z - 3}{ - 2} }=\lambda \: ..(1)$

Direction ratios of l are ( 1,-1,-2)

Given plane p is 2x - y+z - 4= 0

The normal of p is ( 2,-1,1)

Therefore the direction ratios of the required line are m ' =p × l

$\implies \sf{m '= 3 \hat{i} + 5 \hat{j} - \hat{k}}$

Hence , direction ratios of m ' are (3,5,-1)

Now,

From eq .(1)

$\implies \star \sf{ x - 2 = \lambda } \: \: ,\: \: x \: = \lambda \: + 2 \\ \\ \implies \star \sf{ \: y - 2 = - \lambda \: \: \: ,\: y \: = 2 - \lambda \: } \\ \\ \implies \star \sf{z \: - 3 = - 2 \lambda \: \: , \: z = 3 - 2 \lambda} \\ \\ \sf{ \: k ( \lambda + 2 \:, 2 - \lambda \:, \: \: 3 - 2 \lambda \: )}$

Now , substitute these values in the give plane ,

$\rightarrow \small \sf{ 2( \lambda + 2) - (2 - \lambda) + 3 - 2 \lambda - 4 = 0} \\ \\ \rightarrow \sf{2 \lambda + 4 - 2 + \lambda - 2 \lambda - 1 = 0} \\ \\ \rightarrow \boxed{ \lambda \: = - 1}$

Put the value of lambda in K

$\implies \: ( \: 1 \:, \: \:3 \: , \: 5)$

Required equation of line have direction ratios m ' and passing through K ( 1,3,5)

Therefore equation of line is ,

$\rightarrow \boxed{ \sf{ \frac{x - 1}{3} = \frac{y - 3}{5} = \frac{z -5 }{ - 1}} }$