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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 is2(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 ......... |
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Answer» 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 |
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