**23.** The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through

Given equations of planes are
and
Now, from equation (i) and (ii) it is clear that given planes are parallel to each other
Therefore, the correct answer is (B)

**22.** Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units (C) 8 units (D)

Given equations are
and
Now, it is clear from equation (i) and (ii) that given planes are parallel
We know that the distance between two parallel planes is given by
Put the values in this equation
we will get,
Therefore, the correct answer is (D)

**21.** Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then .

The equation of plane having a, b and c intercepts with x, y and z-axis respectively is given by
The distance p of the plane from the origin is given by
Hence proved

**20.** Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

and

Given
Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)
Now the two vectors which are parallel to the two lines are
and
As we know, a vector perpendicular to both vectors and is , so
A vector parallel to this vector is
Now as we know the vector equation of the line which passes through point p and parallel to vector d is
Here in our question, give point p =...

**19. ** Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes and .

Given
A point through which line passes
two plane
And
it can be seen that normals of the planes are
since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.
So, a vector perpendicular to both these normal vector is
Now a line which passes through and parallels to is
So the required line is

**18** Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line and the plane .

Given,
Equation of a line :
Equation of the plane
Let's first find out the point of intersection of line and plane.
putting the value of into the equation of a plane from the equation from line
Now, from the equation, any point p in line is
So the point of intersection is
SO, Now,
The distance between the points (-1,-5,-10) and (2,-1,2) is
Hence the required distance is 13.

**17.** Find the equation of the plane which contains the line of intersection of the planes and which is perpendicular to the plane

The equation of the plane passing through the line of intersection of the given plane in
,,,,,,,,,,,,,(1)
The plane in equation (1) is perpendicular to the plane,
Therefore
Substituting in equation (1), we obtain
.......................(4)
So, this is the vector equation of the required plane.
The Cartesian equation of this plane can be obtained by...

**16.** If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

We have the coordinates of the points and respectively.
Therefore, the direction ratios of OP are
And we know that the equation of the plane passing through the point is
where a,b,c are the direction ratios of normal.
Here, the direction ratios of normal are and and the point P is .
Thus, the equation of the required plane is

**15.** Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.

So, the given planes are:
and
The equation of any plane passing through the line of intersection of these planes is
..............(1)
Its direction ratios are and = 0
The required plane is parallel to the x-axis.
Therefore, its normal is perpendicular to the x-axis.
The direction ratios of the x-axis are 1,0, and 0.
Substituting in equation (1), we obtain
So, the...

**14.** If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane then find the value of p.

Given that the points and are equidistant from the plane
So we can write the position vector through the point is
Similarly, the position vector through the point is
The equation of the given plane is
and We know that the perpendicular distance between a point whose position vector is and the plane, and
Therefore, the distance between the point and the given plane is
...

**13.** Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Given
two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
the normal vectors of these plane are
Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :
Now, as we know
the equation of a plane in vector form is :
Now Since this plane passes through the point (-1,3,2)
Hence the equation of the plane is

**12.** Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

We know that the equation of the line that passes through the points and is given by the relation;
and the line passing through the points, .
And any point on the line is of the form.
This point lies on the plane,
or .
Hence, the coordinates of the required point are or .

**11.** Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

We know that the equation of the line that passes through the points and is given by the relation;
and the line passing through the points,
And any point on the line is of the form .
So, the equation of ZX plane is
Since the line passes through YZ- plane,
we have then,
or and
So, therefore the required point is .

**10.** Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

We know that the equation of the line that passes through the points and is given by the relation;
and the line passing through the points,
And any point on the line is of the form .
So, the equation of the YZ plane is
Since the line passes through YZ- plane,
we have then,
or and
So, therefore the required point is

**9.** Find the shortest distance between lines and .

Given lines are;
and
So, we can find the shortest distance between two lines and by the formula,
...........................(1)
Now, we have from the comparisons of the given equations of lines.
So,
and
Now, substituting all values in equation (3) we get,
Hence the shortest distance between the two given lines is 9 units.

**8.** Find the equation of the plane passing through (a, b, c) and parallel to the plane .

Given that the plane is passing through and is parallel to the plane
So, we have
The position vector of the point is,
and any plane which is parallel to the plane, is of the form,
. .......................(1)
Therefore the equation we get,
Or,
So, now substituting the value of in equation (1), we get
.................(2)
So, this is the required equation...

**7.** Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

Given that the plane is passing through the point so, the position vector of the point A is and perpendicular to the plane whose direction ratios are and the normal vector is
So, the equation of a line passing through a point and perpendicular to the given plane is given by,
, where
.

**6.** If the lines and are perpendicular, find the value of k.

Given both lines are perpendicular so we have the relation;
For the two lines whose direction ratios are known,
We have the direction ratios of the lines, and are and respectively.
Therefore applying the formula,
or
For, the lines are perpendicular.

**5.** If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Direction ratios of AB are
and Direction ratios of CD are
So, it can be noticed that,
Therefore, AB is parallel to CD.
Thus, we can easily say the angle between AB and CD which is either .

**4.** Find the equation of a line parallel to x-axis and passing through the origin.

Equation of a line parallel to the x-axis and passing through the origin is itself x-axis.
So, let A be a point on the x-axis.
Therefore, the coordinates of A are given by , where .
Now, the direction ratios of OA are
So, the equation of OA is given by,
or
Thus, the equation of the line parallel to the x-axis and passing through origin is

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