Question

If the points (1,1,p) and (−3,0,1) are at equidistant from the plane →r.(3ˆi+4ˆj−12ˆk)+13=0, then find the value of p.

Answer 1

Step 1:

Let the position vector through the point (1,1,p)(1,1,p) is a→1=i^+j^+pk^a→1=i^+j^+pk^

Similarly let the position vector through the point (−3,0,1)(−3,0,1) be a→2=−4i^+k^a→2=−4i^+k^

The equation of the given plane is r→.(3i^+4j^−12k^)+13=0r→.(3i^+4j^−12k^)+13=0

The perpendicular distance between a point whose position vector is a→a→ and the plane r→.N→=dr→.N→=d,is given by D=∣∣∣∣a→.N→−d→∣N→∣∣∣∣∣D=|a→.N→−d→∣N→∣|

Here N→=3i^+4j^−12k^N→=3i^+4j^−12k^ and d=−13d=−13.

Step 2:

Now substituting this we get

D1=∣(i^+j^+pk^).(3i^+4j^−12k^)+13∣∣3i^+4j^−12k^∣D1=∣(i^+j^+pk^).(3i^+4j^−12k^)+13∣∣3i^+4j^−12k^∣

We know that i^.i^=j^.j^=k^.k^=1i^.i^=j^.j^=k^.k^=1

D1=∣3+4−12p+13∣32+42−122−−−−−−−−−√D1=∣3+4−12p+13∣32+42−122

=20−12p13=20−12p13

Step 3:

Similarly the distance between the point (−3,0,1)(−3,0,1) and the given plane is

D2=∣(−3i^+k^).(3i^+4j^−12k^)+13∣32+42+(−12)2−−−−−−−−−−−−√D2=∣(−3i^+k^).(3i^+4j^−12k^)+13∣32+42+(−12)2

D2=813D2=813

Step 4:

But it is given that the distance between the required plane and the points (1,1,p)(1,1,p) and (−3,0,1)(−3,0,1) equal.

Therefore D1=D2D1=D2

Hence 20−12p13=81320−12p13=813

Step 5:

On simplifying we get

∣20−12p∣=8∣20−12p∣=8

(i.e) 20−12p=820−12p=8

Case (i)⇒−12p=−12⇒−12p=−12

p=1p=1

Case (ii)⇒−20+12p=8⇒−20+12p=8

−20+12p=8−20+12p=8

12p=2812p=28

p=2812=73p=2812=73

Therefore p=1p=1 or 73