Vector & 3D Geometry (II)

The direction cosines of the x axis are given by

• (1,0,0)

• (0,0,1)

• (2,0,1

• (0,1,0)

Any point on the x axis have coordinates of the form

• (a,0,0)

• (0,a,a)

• (0,O,a)

• (0,a,0)

A line is equally inclined to the x,y and z axis ,the direction cosines of the line are given by

• 1,1,1

• 1,2,1

• 1/2,1/2,1/2

Perpendicular drawn from a point (6,5,4) to the x axis meets it at the point

• (6,0,4)

• (4,0,0

• (6,0,1)

• (6,0,0)

The point ( -1 , 2 , -3 ) lies in the :

• (A). 7th Octant

• (B). 5th Octant

• (C). 6th Octant

• (D). 8th Octant

Find the equation of the line which passes through the point (1, 2, and 3) and is parallel to the vector 3î+2ĵ -2k̂.

Answer:-

• r = î +2ĵ +3k̂+λ (3î +2ĵ -2k̂).

• r = î +2ĵ +3k̂+λ (3î -2ĵ -2k̂).

• r = î +2ĵ -3k̂+λ (3î +2ĵ -2k̂).

• r = î +2ĵ +3k̂+λ (3î +2ĵ -3k̂).

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î-ĵ +4k̂ and is in the direction î +2ĵ -k̂?

• r =2î -ĵ +4k̂+ λ(2î +2ĵ -k̂)

• r =2î -ĵ +4k̂+ λ(î +2ĵ -k̂)

• r =2î +ĵ +4k̂+ λ(î +2ĵ -k̂)

• r =2î -2ĵ +4k̂+ λ(î +2ĵ -k̂)

Find the Cartesian equation of the line which passes through the point(−2, 4, −5) and parallel to the line given by (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)?

• (x+2)/ (3k) =(y-4)/ (5k) = (z+5)/ (6k)

• (x+2)/ (3k) =(y-4)/ (5k) = (z-5)/ (6k)

• (x+2)/ (3k) =(y+4)/ (5k) = (z+5)/ (6k)

• (x-2)/ (3k) =(y-4)/ (5k) = (z+5)/ (6k)

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

• r = (3î -2ĵ -5k̂) +12λk̂

• r = (3î -2ĵ +5k̂) +11λk̂

• r = (3î +2ĵ -5k̂) +11λk̂

• r = (3î -2ĵ -5k̂) +11λk̂

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3)

• r->=λ(5î -3ĵ +3k̂)

• r->=λ(5î -2ĵ -3k̂)

• r =λ(5î -2ĵ +3k̂)

• r->=λ(5î +2ĵ +3k̂)