Question

solve this question.

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Answer 1

Solution:-

Let P be the point on lines.

(2λ+1,-λ-1 ,λ) foot of perpendicular

Q is given by

x-2λ-1/1 = y+λ+1/1 = z-λ/1 = -(2λ-3)/1

∴ Q lies on x+y+z= 3 and x-y+z= 3

⇒x+z= 3 and y= 0

y=0⇒λ +1= -2λ +3/3⇒λ= 0

⇒Q is ( 2,0,1)

above other method...

Hope it helps you..

Answer 2

$\Large\underline\mathfrak{Question}$

• Given above in image ...
• we have to find co-ordinates of Q .

$\Large\underline{\underline{\sf{Solution}:}}$

$\red{\textbf{Refer To image}}$

Steps :-----

1) let the Equation is Equal to lem. (opp. of Y)

2) foot of perp. is :----

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{ - (ab_1 + by_1 + cz_1)}{a + b + c}$

3) putting values here , and than comparing , first we find foot of perp.

4) it is also given that this perp. also lies on plane (x-y+z) = 3 .

5) so, we find value of opp. y . and putting this we find value of perp..

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$\large\underline\textbf{Hope it Helps You.}$