Question:

If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane  r(3i^+4j^12k^)+13=0,\vec r(3\hat i+4\hat j-12\hat k)+13=0, then find the value of p.

Updated On: Sep 21, 2023
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Solution and Explanation

The position vector through the point (1, 1, p) is
a1=i^+j^+pk^\vec a_1 =\hat i+\hat j+p\hat k

Similarly, the position vector through the point (-3, 0, 1)is
a2=4i^+k^\vec a_2 =-4\hat i+\hat k

The equation of the given plane is
r.(3i^+4j^12k^)+13=0\vec r.(3\hat i+4\hat j-12\hat k)+13=0

It is known that the perpendicular distance between a point whose position vector is a\vec a and the plane,r.N=d\vec r.\vec N =d, is given by,D=a.NdND=\frac {|\vec a.\vec N-d|}{|\vec N|}

Here, N=3i^+4j^12k^\vec N=3\hat i+4\hat j-12\hat k and d=13d=-13

Therefore,the distance between the point(1,1,p)and the given plane is

D1=(i^+j^+pk^).(3i^+4j^12k^)+133i^+4j^12k^D_1=\frac {|(\hat i+\hat j+p\hat k).(3\hat i+4\hat j-12\hat k)+13|}{|3\hat i+4\hat j-12\hat k|}

D1=3+412p+1332+42+(12)2D_1=\frac {|3+4-12p+13|}{\sqrt {3^2+4^2+(-12)^2}}

D1=2012p13D_1=\frac {|20-12p|}{13}        ...(1)

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

D2=(3i^+k^).(3i^+4j^12k^)+133i^+4j^12k^D_2=\frac {|(-3\hat i+\hat k).(3\hat i+4\hat j-12\hat k)+13|}{|3\hat i+4\hat j-12\hat k|}

D2=912+1332+42+(12)2D_2=\frac {|-9-12+13|}{\sqrt {3^2+4^2+(-12)^2}}]

D2=813D_2=\frac {8}{13}        ...(2)

It is given that the distance between the required plane and the points (1, 1, p) and (-3, 0, 1) is equal.

D1=D2∴D_1=D_2

⇒ 2012p13=813\frac {|20-12p|}{13} =\frac {8}{13}

⇒ 2012p=8 or (2012p)=820-12p=8 \ or\ -(20-12p)=8

⇒ 12p=12 or 12p=2812p=12 \ or \ 12p=28

⇒ p=1  or p=73p=1\  or \ p=\frac 73

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