Question

Fill in the blanks :
(i) The perpendicular distance of a point (5, 12, 13) from Y-axis is ...
(ii) The equation of the plane passing through the origin (0,0,0) is ....
(iii) The intercept cut by the plane 2x + y - z=5 on X-axis is
2
(iv) The centre of the sphere x² + y2 +2? +3x + 5y +22=0 is
(v) The differential coefficient of sin 3x with respect to 3x is .........​

Answer 1

Answer:

The equation of any plane passing through the lines of intersection of the given planes

x+2y+3z−5=0 and 3x−2y−z+1=0 is

(x+2y+3z−5)+k(x−2y−z+1)=0 ....... (i)

For the intercept on x-axis, on putting y = 0 and z = 0, we get

x+3kx−5+k=0⇒x=

3k+1

5−k

For the intercept on z-axis, on putting x = 0, y = 0, we get

3z−kz−5+k=0⇒z=

3−k

5−k

∴ Intercepts on x-axis and z-axis made by the plane (i) are

3k+1

5−k

and

3−k

5−k

respectively.

Since the intercepts on x-axis and z-axis are equal.

3k+1

5−k

=

3−k

5−k

⇒−3k

2

+14k+5=−8k+k

2

+15

⇒4k

2

+22k−10=0

⇒4k

2

−22k+10=0

⇒(4k−2)(k−5)=0

⇒k=

2

1

,5

On putting k=5 in (i), we notice that the plane passes through origin and hence, it cannot make intercepts on axes. Therefore k=

2

1

is the only admissible value.

Substituting k=

2

1

in equation (i) the equation of the required plane is

x+2y+3z−5+2/2(3x−2y−z+1)=0

⇒2x+4y+6z−10+3x−2y−z+1=0

⇒5x+2y+5z−9=0