Question

QUESTION
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If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane vector r.(3i+4j 12k)+13=0 ,then find the value of p.





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Harsh Pratap Singh ​

Answer 1

S O L U T I O N : -

ˉ (−3 i^ −4 j^ +12 k^ )=13

Comparing with rˉ . nˉ =d d=13 Magnitude of n ˉ = (−3) 2 +(−4) 2 +12 2

⇒∣ nˉ ∣= 9+16+144

= 169

=13

Distance of a 1 ˉ from plane

= ∣∣∣∣∣∣∣ nˉ ∣a 1ˉ . nˉ −d ∣∣∣∣∣∣

= ∣∣∣∣∣∣∣ 13( i^ + j^ +p k^ )(−3 i^ −4 j^ +12 k^ )−13 ∣∣∣∣∣∣∣

= ∣∣∣I∣∣ 13−3−4+12p−13 ∣∣∣∣∣∣

= ∣∣∣∣∣∣13 12p−20 ∣∣∣∣∣∣ ⟶1

Distance of pont a 2ˉ from plane

= ∣∣∣∣l∣∣ nˉ ∣a 2ˉ . nˉ−d ∣∣∣∣∣∣

= ∣∣∣∣∣∣∣ 13(−3 i^ +0 j^ + k^ )(−3 i^ −4 j^ +12 k^ )−13 ∣∣∣∣∣∣∣

= ∣∣∣∣∣∣ 13−9+0+12−13 ∣∣∣∣∣∣

= ∣∣∣∣∣∣ 138 ∣∣∣∣∣∣

= 138 ⟶2

∵ a1 ˉ and a 2

ˉ are equidistance from plane

⇒1 = 2

⇒∣12p − 20∣ = 8

⇒12p − 20 = ± 8

⇒12p = 28

⇒p = 37

⇒12p − 20 = −8

⇒p = 12

12 = 1

∴p = 1, 37

hope it's help