Math, asked by muskanrathore2271, 1 year ago

A perpendicular is drawn from a point on the line (x - 1)/2 = (y - 1)/-1 = z/1 to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are
(A) (2, 0, 1) (B) (-1, 0, 4)
(C) (1, 0, 2) (D) (4, 0, -1)

Answers

Answered by yogichaudhary
0

\huge\boxed{\fcolorbox{black}{pink}{Hi mate!}}

A perpendicular is drawn from a point on the line (x - 1)/2 = (y - 1)/-1 = z/1 to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are

(A) (2, 0, 1)

(B) (-1, 0, 4)✔

(C) (1, 0, 2)

(D) (4, 0, -1)

Answered by ʙʀᴀɪɴʟʏᴡɪᴛᴄh
1

Answer:

\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}

A perpendicular is drawn from a point on the line (x - 1)/2 = (y - 1)/-1 = z/1 to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are

(A) (2, 0, 1) ...................(B) (-1, 0, 4)

(C) (1, 0, 2) (D) (4, 0, -1)

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