Question

# Consider a plane $$x+y-z=1$$ and point $$A(1, 2, -3)$$. A line $$L$$ has the equation $$x=1+3r, y=2-r$$ and $$z=3+4r$$.The coordinate of a point $$B$$ of line $$L$$ such that $$AB$$ is parallel to the plane is:

A
(10,1,15)
B
(5,4,5)
C
(4,1,7)
D
(8,5,9)

Solution

## The correct option is D $$(-8, 5, -9)$$Let $$\vec { OB } =\left( 1+3r \right) i+\left( 2-r \right) j+\left( 3+4r \right) k$$$$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right) i+\left( 2-r \right) j+\left( 3+4r \right) k-i-2j+3k=3ri-rj+\left( 6+4r \right) k$$Since, $$\vec { AB }$$ is parallel to $$x+y-z=1$$Therefore, $$\vec { AB } .\left( i+j-k \right) =0$$$$\Rightarrow \left( 3ri-rj+\left( 6+4r \right) k \right) .\left( i+j-k \right)$$$$\Rightarrow 3r-r-6-4r=0$$$$\Rightarrow r=-3$$Therefore, $$\vec { OB } =-8i+5j-9k$$Ans: DMaths

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