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 .........
Answers
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