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51.

Find the equation of the plane which passes through the point (2,-3,7) and makes equal intercepts on the coordinate axes.

Answer» Correct Answer - `x+y+z=6`
Let it make intercept a on each of the coordinate axes. Then, its equation is `x/a+y/a+z/a=1 rArr x+y+z=a`.
Putting, `x=2,y=-3,z=7`, we get `a=2+(-3)+7=6`.
So, the required equation of the plane is `x+y+z=6`.
Discussion
52.

Find the equation of the plane which cuts off intercepts 3,6 and -4 from the axes of coordinates.

Answer» We know that the equation of a plane which cuts off intercept a,b,c from the x-axis, y-axis and z-axis respectively, is
`x/a+y/b+z/c=1`.
Here, a=3, b=6 and c=-4.
Hence, the required equation of the plane is
`x/3+y/6+z/-4=1 rArr 4x+2y-3z=12`.
Discussion
53.

Write the equation of the plane whose intercepts on the coordinate axes are 2,-4 and 5 respectively.

Answer» Correct Answer - `10x-5y+4z=20`
Discussion
54.

A variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axes is `(1//2)`. Then, the plane passes through the pointA. `(0,0,0)`B. (1,1,1)C. `(1/2,1/2,1/2)`D. `(2,2,2)`

Answer» Correct Answer - d
Let the required equation of the plane be `x/a+y/b+z/c=1`.
Then, `1/a+1/b+1/c =1/2 rArr 2/a+2/b+2/c=1`.
It means that the plane passes through the point (2,2,2).
Discussion
55.

Reduce the equation `2x-3y+5z+4=0` to intercept form and find the intercepts made by it on the coordinate axes.

Answer» Correct Answer - `-2,4/3,-4/5`
The given equation may be written as `-2x+3y-5z=4`.
`rArr (-2x)/4+(3y)/4+)(-5z)/4=1 rArr x/-1+y/(4/3)+z/(-4/5)=1`.
`therefore the required intercepts are `-2,4/3,-4/5`.
Discussion
56.

The line `(x-4)/1=(y-2)/1=(z-k)/2` lies exactly on the plane `2x=4y+z=7` then the value of k is (A) 7 (B) -7 (C) 1 (D) none of theseA. `-7`B. 7C. 4D. `-4`

Answer» Correct Answer - b
Clearly, the given line passes through the point (4,2,k).
Since, the given line lies in the plane `2x-4y+z=7`, so the above point lies in this plane.
`therefore (2xx4)-(4 xx 2)+k=7 rArr k=7`.
Discussion
57.

A plane cuts off intercepts `3,-4,6` on the coordinate axes. The length of perpendicular from the origin to this plane isA. `5/sqrt(29)` unitsB. `8/sqrt(29)` unitsC. `6/sqrt(29)` unitsD. `12/sqrt(29)` units

Answer» Correct Answer - d
The given plane is `x/3+y/-4+z/6=1 rArr 4x-3y+2z=12`
`therefore p=(|4 xx 0-3 xx0+2 xx 0-12|)/sqrt(4^(2)+(-3)^(2)+2^(2))=12/sqrt(29)` units.
Discussion
58.

Find the equation of a plane passes through the point `(0 ,0,0)` and perpendicular to each to the planes `x+2y-z=1` and `3x-4y+z=5`.

Answer» Correct Answer - `x+2y+5z=0`
Discussion
59.

Find the value of `lambda` for which the given planes are perpendicular to the each other: i) `vecr.(2hati-hatj-lambdak)=7` and `vecr.(3hati+2hatj+2hatk)=9` ii) `vecr.(lambdahati+2hatj+3hatk)=5` and `vecr.(hati+2hatj-7hatk)+11=0`

Answer» Correct Answer - i) `lambda=2`, ii) `lambda=17`
Discussion
60.

Find the equation of the plane passing through the points A(1,-1,2) and B(2,-2,2) and perpendicular to the plane `6x-2y+2z=9`.

Answer» Correct Answer - `x+y-2z+4=0`
Discussion
61.

Find the Cartesian from the equation of the plane `vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk`.

Answer» We have,
`vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk`
`ltimplies (xhati+yhatj+zhatk)=(s-2t)hati+(3-t)hatj+(2s+t)hatk`
`ltimplies x-2y+6=1/2(y+z-3)` [equating the values of x]
`ltimplies 2x-4y+12=y+z-3 rArr 2x-5y-z+15=0`.
This is the required Cartesian form of the equation of the given plane.
Discussion
62.

The equation of a plane passing throgh the points `A(a,0,0), B(0,b,0)` and C(0,0,c) is given byA. `ax+by+cz=0`B. `ax+by+cz=1`C. `x/a+y/b+z/c=0`D. `x/a+y/b+z/c=1`

Answer» Correct Answer - d
The required equation is `x/a+b/b+z/c=1`.
Discussion
63.

Find the equation of a plane containing the line of intersectionof the planes `x+y+z-6=0a n d2x+3y+4z+5=0`passing through `(1,1,1)`.

Answer» Correct Answer - `20x+23y+26z=69`
Discussion
64.

Find the equation of the plane passing through the point `(2,-3,5)` and parallel to the points `3x-7y-2z=5`. Also, the find the distance between the two planes.

Answer» The given plane is `3x-7y-2z-5=0`……………..(i)
Let the required plane parallel to the given plane be
`3x-7y-2z+lambda=0` for some real number `lambda`…………(ii)
Since, the plane (ii) passes through the point `(2,-3,5)` , we have
`(3xx3)-7 xx (-3)-(2xx5)+lambda=0 rArr (6+21-10)+lambda=0`
`rArr lambda=-17`.
`therefore` the equation of the required plane is `(3x-7y-2z-17=0`...........(iii)
Now, the planes (i) and (ii) are parallel and the point (2,-3,5) lies on (iii).
`therefore` distance between the two planes.
= distance between the point (2,-3,5) and the plane (i)
`=(|(3xx2)-7xx(-3)-(2xx5)-5|)/(sqrt(3^(2)+(-7)^(2)+(-2)^(2))`
`=(|6+21-10-5|)/(sqrt(9+49+4))`
`=12/sqrt(62) xx sqrt(62)/sqrt(62) = 12/62sqrt(62)`
`=6/31sqrt(62)` units.
Hence, the distance between the two planes is `6/31 sqrt(62)` units.
Discussion
65.

The plane passing through the point (4, -1, 2) and perallel to the lines `(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)` also passes through the point

Answer» The general equation of a plane passing throught the point A(4,-1,2) is given by
a(x-4) +b(y+1)+c(z-2)=0
This plane will be parallel to each of the given lines only when the normal to the plane is perpendicuale to each of the given lines
`therefore (1xxa)+(2xxb)+(3xxc)=0 rarr a+2b+3c=0`
`(3xxa)+(-1)xxb+)=0 rarr3a-b+2c=0`
on solving (ii) and (iii) by cross multiplication we get
`rarr (a)/(7) =(b)/(7)=(c )/(-7) rarr (a)/(1)=(b)/(1)=(c )/(-1)=lambda`(say)
`rarr a=lambda , b = lambda` and `c=-lambda`
Putting these values of a,b c in (i) we get
`lambda (x-4) + lambda(y+1) -lamdba(z-2)=0`
`rarr (x-4) +(y+1)-(z-2)=0 rarr x+y -z=1`
`Hence the required equation of the plane is x+y-z=1
Discussion
66.

Find the equation of a plane passing through the points A(a,0,0), B(0,b,0) and C(0,0,c)`.

Answer» Correct Answer - `x/a+y/b+z/c=1`
Clearly,the required equation of the planes is `x/a+y/b+z/c=1`.
Discussion
67.

Find the vector equation of the plane passing thrugh the points (2,5,-3),(-2,-3,5),(5,3,-3).

Answer» The equation of the plane passing through the point A(2,5,-3) is given by
`a(x-2)+b(y-5)+c(z+3)=0`.
If the plane passes through the points B(-2,,-3,5) and C(5,3,-3), then we have
`a(-2-2)+b(-3-5)+c(5+3)=0 rArr -4a-8b+8c=0`……………(i)
If the plane passes through the points B(-2,-3,5) and C(5,3,-3), then we have
`a(-2-2)+b(-3-3)+c(5+3)=0 rArr -4a-8b+8c=0.............(ii)`
`a(5-2)+b(3-5)+b(3-5)+c(-3+3)=0 rArr 3a-2b+0c=0`..................(iii)
On solving (ii) and (iii) by cross multiplication, we get
`a/(0+16)=b/(24-0)=c/(8+24) rArr a/16=b/24=c/32`
`rArr a/2=b/3=c/4=k` (say) `rArr a=2k, b=3k` and `c=4k`.
Putting these values of a,b and c in (i), we get
`2k(x-2)+3k(y-5)+4k(z+3)=0`
`rArr 2(x-2)+3(y-5)+4(z+3)=0`
`rArr 2(x-2)+3(y-5)+4(z+3)=0`
`rArr 2x+3y+4z-7=0`.
Hence, the required equation of the plane is `2x+3y+4z-7=0`.
Discussion