The projection of the line `(x+1)/(-1)=y/2=(z-1)/3` on the plane `x-2y+z=6` is the line of intersection of this plane with the plane a. `2x+y+2=0` b.

Question

The projection of the line `(x+1)/(-1)=y/2=(z-1)/3` on the plane `x-2y+z=6` is the line of intersection of this plane with the plane a. `2x+y+2=0` b.

The projection of the line `(x+1)/(-1)=y/2=(z-1)/3` on the plane `x-2y+z=6` is the line of intersection of this plane with the plane a. `2x+y+2=0` b. `3x+y-z=2` c. `2x-3y+8z=3` d. none of these
A. `2x+y+2=0`
B. `3x+y-z=2`
C. `2x-3y+8z=3`
D. none of these

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - a
Equation of the plane through `(-1, 0, 1)` is
`" "a(x+1)+b(y-0)+c(z-1)=0" "` (i)
Which is parallel to the given line and perependicular to the given plane
`" "-a+2b+3c=0" "`(ii)
and `" "a-2b+c=0" "`(iii)
From Eq. (i) and (iii), we get
`" "c=0, a= 2b`
From Eq. (i), 2b (x+1)+ by =0
`rArr" "2x+y+2=0`

5 views Answered 2023-02-05 15:29