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1.

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

Answer» Correct Answer - A
Equation of line parallel to given line and through `(1,-2,3)` is `(x-1)/2=(y+2)/3=(z-3)/-6` …………(1)
Discussion
2.

The locus represented by `xy+yz=0` is a pair ofA. perpendicular linesB. parallel linesC. parallel linesD. perpendicular planes

Answer» Correct Answer - D
`xy+yz=0`
`therefore y(x+z)=0`
`therefore y=0 or x+z=0`
This is a pair of perpendicular planes.
Discussion
3.

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`

Answer» Correct Answer - D
The plane is `x+3y+z-4+lambda(2x-y)=0`
This is always passes through the line of intersection of the planes `x+3y+z=4=0` and `2x-y=0`
Now, `2x-y=0 rArr x/1=y/2`
`therefore x+3y+z-4=0`
`rArr x+3.2x+z-4=0`
`rArr x+3.2x+z-4=0`
`rArr 7x+z-4=0`
`rArr 7x = -(z-4)`
`rArr x/1=(z-4)/-7`
`therefore` line is `x/1=y/2=(z-4)/-7`
Discussion
4.

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
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`

Answer» Correct Answer - B
`vecn_(1) = 3hati-3hatj+hatk` and `vecn_(2) = hati-hatj+chatk`
`cos 45^(@) = |(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)|=|((3+3+c)/(sqrt(19)sqrt(2+c^(2))))|`
`therefore 2(6+c)^(2) = 19(2+c^(2))`
`rArr 2(36+c^(2)+12c)=38+19c^(2)`
`rArr 17c^(2)-24c-34=0`
Discussion
6.

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

Answer» Correct Answer - B
Volume `[bar(AB)bar(AC)bar(AD)]=|{:(2,3,-2),(6,-4,-8),(0,2,-5):}|`=138
Discussion
7.

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

Answer» Correct Answer - A
Line L is `(x-1)/1=(y-2)/2sqrt(2)=z/0`
This line L makes an angle of `45^(@)` with the plane `sqrt(2)x+y-z=1`
`therefore` Required distance PQ is prependicular distance of plane from P
I.e., `PQ=(|sqrt(2)+7-sqrt(2)-1|)/sqrt(2+1+1)=3`
Discussion
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

Answer» Correct Answer - B::C::D
`E_(1): vecr xx (2hati-hatj+3hatk)=3hati+hatk`
`rArr 3hati(3y+z)-hatj(3x-2z)+hatk(-x-2y)=3hati+hatk`
`therefore 3y+z=3, 3x-2y=0, -x-2y=1`
From first two equations, `3x-2(3-3y)=0`
`rArr 3x+6y=6 rArr x+2y=2`
Now, `x+2y=-1, x+2y=2` are parallel planes.
`E_(2): vecr xx (hati+2hatj-3hatk)=2hati-hatj`
`rArr hati(-3y-2z)-hatj(-3x-z)+hatk(2x-y)=2hati-hatj`
`therefore hati(-3y-2z)-hatj(-3x-z)+hatk(2x-y)=2hati-hatj`
`therefore -3y-2z=2, 3x+z=-1, 2x-y=0`
i.e., `-6x-2z=2, 3x+z=-1`
`therefore` Straight line, `2x-y=0, 3x+z=-1`
Discussion
9.

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`

Answer» Correct Answer - D
We have `Q(-1,0,1)` and R(1,0,-1).
`therefore QR=2sqrt(2)`
Now, the mid point of QR is M(0,0,0).
Triangle PQR is equilateral.
`therefore PM=MQ xx tan60^(@) = sqrt(2)sqrt(3)=sqrt(6)`.
Thus, P can lie on the plane whose distance from point M is less than or equal to `sqrt(6)`.
Hence, P can lie on the plane `x+y+z+3sqrt(2)=0`
Discussion
10.

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

Answer» Correct Answer - A
Any plane passing through first line is `2x+y=z-1+lambda(3x+y+2z-2)=0`
Line x=y=z is passing through the point O(0,0,).
Required shortest distance = distance of O from the member plane of above family which is parallel to the line x=y=z
If plane is parallel to the line,
`(2x+3lambda)1+(1+lambda)1+(1+2lambda)1=0`
`rArr lambda=-2/3`
Equation of plane is
`3(2x+y+z-1)-2(3x+y+2z-2)=0`
or `y-z+1=0`
Its distance from (0,0,0) is `1/sqrt(2)`.
Discussion