Related InterviewSolutions

• The equaton of the line in vector and cartesian from that passes through the point with position vector ` 2 hati - hatj + 4hatk ` and is in the direction ` hati + 2 hatj - hatk ` areA. ` r = ( 2hati - hatj + 4hatk ) + lamda ( hati + 2 hatj - hatk ) , ( x - 1 ) / ( 2 ) = ( y - 2 ) / ( - 1 ) = ( z + 1 ) / ( - 4 ) `B. ` r = ( hati + 2 hatj - hatk ) + lamda ( 2 hati - hatj + 4 hatk ) , ( x - 1 ) / ( 2 ) = ( y - 2 ) / ( - 1 ) = ( z + 1 ) / (4 ) `C. ` r = ( 2hati - hatj + 4 hatk ) + lamda ( hati + 2 hatj - hatk ), ( x - 2 ) / ( 1 ) = ( y + 1 ) / ( 2 ) = ( z - 4 ) / ( - 1 ) `D.
• Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. (9,5,)B. (9,-5,3)C. (9,5,-3)D. (-9,5,3)
• The cartesian equation of a line passing through the point having position vector `2hati +hatj - hatk` and parallel to the line passing joining the points `-hati + hatj + 4hatk` and `hati + 2hatj + 2hatk`, isA. `(x-2)/2 = (y-1)/1 = (z+1)/-2`B. `(x+2)/2 = (y+1)/1 = (z-1)/-2`C. `(x-2)/-2 = (y-1)/1 = (z+1)/2`D. `(x+2)/-2 = (y+1)/1 = (z-2)/2`
• Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. 3B. 2C. 7D. 6
• Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. `sqrt(531/14)`B. `sqrt(531/98)`C. `sqrt(1779/98)`D. `sqrt(1779/49)`
• The equation of a line passing through (a, b, c) and parallel tp z-axis isA. `(x-a)/l - (y-b)/m = (z-c)/0`B. `(x-a)/0 = (y-b)/0 =(z-c)/n`C. `(x-a)/l = (y-b)/m = (z-c)/1`D. `(x-a)/0 = (y-b)/m = (z-c)/0`
• The foot of the perpendicular drawn from point (2,-3,1) to the line `(x+1)/2 = (y-3)/3 = (z+2)/-1` isA. `(-22/14, -3/14, -13/14)`B. `(-22/7, -3/14, -13/4)`C. `(22/14, 3/14, 13/14)`D. `(22/7, 3/14, 13/14)`
• The equation of a line passing through point (2,3,-1) and perpendicular to XY plane isA. `(x-2)/0 = (y-3)/1 = (z+1)/0`B. `(x-2)/1 = (y-3)/0 = (z+1)/0`C. `(x-2)/0 = (y-3)/0 = (z+1)/1`D. `(x-2)/1 = (y-3)/1 = (z+1)/1`
• Find the equation of the perpendicular from point `(3,-1,11)` to line `x/2=(y-2)/3=(z-3)/4`. Also, find the coordinates of foot of perpendicular and the length of perpendicular.A. `(x+1)/1 = (y-3)/-6 = (z+11)/3`B. `(x-3)/2 = (y-1)/-6 = (z+11)/3`C. `(x-3)/1 = (y+1)/-6 = (z-11)/4`D. `(x-3)/2 = (y+1)/6 = (z-11)/7`
• Find the equation of the line passing through the point (3,1,2) and perpendicular to the lines `(x-1)/1=(y-2)/2=(z-3)/3` and `x/(-3)=y/2=z/5`A. `(x-3)/2 = (y-1)/-7 = (z-2)/2`B. `(x+3)/2 = (y+1)/7 = (z-2)/2`C. `(x-3)/2 = (y-1)/-7 = (z-2)/4`D. `(x+3)/2 = (y+1)/-7 = (z+2)/4`