Question

Find the distance of the point 2,12,5 from the point of intersection of line r

Answer 1

Equation of the given line is

r→=2i^−4j^+2k^+λ(3i^+4j^+2k^)r→=2i^−4j^+2k^+λ(3i^+4j^+2k^) -----------(1)

Given equation of the plane is

r→=(i^−2j^+k^)=0r→=(i^−2j^+k^)=0--------(2)

Substituting the value of r→r→ in eqn 2 we get

(2i^−4j^+2k^)+λ(3i^+4j^+2k^)=(i^−2j^+k^)=0(2i^−4j^+2k^)+λ(3i^+4j^+2k^)=(i^−2j^+k^)=0

⇒[(2+3λ)i^+(4λ−4)j^+(2+2λ)k^].[i^−2j^+k^]=0⇒[(2+3λ)i^+(4λ−4)j^+(2+2λ)k^].[i^−2j^+k^]=0

⇒(2+3λ)(1)+(4λ−4)(−2)+(2+2λ)(1)=0⇒(2+3λ)(1)+(4λ−4)(−2)+(2+2λ)(1)=0

2+3λ−8λ+8+2+2λ=02+3λ−8λ+8+2+2λ=0Hence the equation of the line is

r→=(2i^−4j^+2k^)+4(3i^+4j^+2k^)r→=(2i^−4j^+2k^)+4(3i^+4j^+2k^)

(i.e.,) r→=2i^−4j^+2k^+12i^+16j^+8k^r→=2i^−4j^+2k^+12i^+16j^+8k^

=14i^+18k^+10k^=14i^+18k^+10k^

The point of intersection of the given line and the plane is given by the co-ordinate

(14, 12, 10), the given point is ( 2, 12, 5)

hence the distance between the point is

d=(14−2)2+(12−12)2+(10−5)2−−−−−−−−−−−−−−−−−−√d=(14−2)2+(12−12)2+(10−5)2

=122+0+52−−−−−−−−−−√122+0+52

=144+25−−−−−−−√144+25

=144+25−−−−−−−√144+25

=169−−−√169

=13=13

hence the distance is 13 units