The Cartesian equation of the plane `vecr=(1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk` isA. `2x+y=5`B. `2x-y=5`C. `2x+z=5`D. `2x-z=5`

The Cartesian equation of the plane `vecr=(1+lambda-mu)hati+(2-lambda)

1.	The Cartesian equation of the plane `vecr=(1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk` isA. `2x+y=5`B. `2x-y=5`C. `2x+z=5`D. `2x-z=5`
Answer» Correct Answer - C We have `vecr = (1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk` `rArr vecr=(hati+2hatj+3hatk)+lambda(hati-hatj-2hatk)+mu(-hati+2hatk)` This is a plane passing through `veca = hati+2hatj+3hatk` and parallel to the vectors `vecb=hati-hatj-2hatk` and `vecc=-hati+2hatk` Therefore, it is perpendicular to the vector `vecn=vecb xx vecc = -2hati-hatk` Hence, its vector equation is `(vecr-veca)=veca.vecn` `rArr vecr. (-2hati-hatk)=-2-3` `rArr vecr.(2hati+hatk)=5` So, the Cartesian equation is `(xhati+yhatj+zhatk).(2hati+hatk)=5` or `2x+z=5`

Discussion

No Comment Found

Related InterviewSolutions

The distance of the point `(1,-2,3)` from the plane `x-y+z-5=0`, measured parallel to the line `x/2=y/3=(z-1)/-6` is equal toA. 1 unitB. 2 unitC. 3 unitsD. none of these
The locus represented by `xy+yz=0` is a pair ofA. perpendicular linesB. parallel linesC. parallel linesD. perpendicular planes
The variable plane `(2lambda+1)x+(3-lambda)y+z=4` always passes through the lineA. `x/0=y/0=(z-4)/1`B. `x/1=y/2=(z-4)/-3`C. `x/1=y/1=(z-4)/-7`D. `x/1=y/2=(z-4)/-7`
The Cartesian equation of the plane `vecr=(1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk` isA. `2x+y=5`B. `2x-y=5`C. `2x+z=5`D. `2x-z=5`
Let P denotes the plane consisting of all points that are equidistant from the points `A(-4,2,1)` and `B(2,-4,3)` and Q be the plane, `x-y+cz=1` where `c in R`. If the angle between the planes P and Q is `45^(@)` then the product of all possible values of c isA. `-17`B. `-2`C. 17D. `24//27`
A line `L_(1)` with direction ratios `-3,2,4` passes through the point A(7,6,2) and a line `L_(2)` with directions ratios 2,1,3 passes through the point B(5,3,4). A line `L_(3)` with direction ratios `2,-2,-1` intersects `L_(1)` and `L_(3)` at C and D, resectively. The volume of parallelopiped formed by `vec(AB), vec(AC)` and `vec(AD)` is equal toA. 140B. 138C. 134D. 130
Let `P=-(1,7,sqrt(2))` be a point and line L is `2sqrt(2)(x-1)=y-2,z=0`. If PQ is the distance of plane `sqrt(2)x+y-z=1` from point P measured along a line inclined at an angle of `45^(@)` with the line L and is minimum then the value of PQ isA. 3B. 4C. 6D. 8
Consider the equation `E_(1):vecr xx (2hati-hatj+3hatk)=3hati+hatk` and `E_(2): vecr xx (hati+2hatj-3hatk)=2hati-hatj`, htenA. `E_(1)` represents a lineB. `E_(1)` represents two parallel linesC. `E_(2)` represents a lineD. `E_(2)` represents two parallel planes
If `P(alpha,beta,lambda)` is a vertex of an equilateral triangle PQR where vertex Q and R are `(-1,0,1)` and `(1,0,-1)` respectively, then P can lie on the planeA. `x+y+z+6=0`B. `2x+4y+3z+20=0`C. `x-y+z+12=0`D. `x+y+z+3sqrt(2)=0`
The shortest distance between the lines `2x+y+z-1=0=3x+y+2z-2` and `x=y=z`, isA. `1/sqrt(2)` unitsB. `sqrt(2)` unitsC. `3/sqrt(2)` unitsD. `sqrt(3)/2` units

The Cartesian equation of the plane `vecr=(1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk` isA. `2x+y=5`B. `2x-y=5`C. `2x+z=5`D. `2x-z=5`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment