Find the vector equation of the plane passing through the intersection of the planes ` -> rdot( hat i+ hat j+ hat k)=6`and ` -> rdot(2 hat i+3 hat j+4 hat k)=-5`and the point (1, 1, 1).

Options

A. `r*(3hat(i)+4hat(j)+5hat(k))=1`
B. `r*(8hat(i)+5hat(j)+2hat(k))=99`
C. `r*(20hat(i)+23hat(j)+26hat(k))=69`
D. `r*(-20hat(i)-23hat(j)-26hat(k))=69`

Answer

C

Explanation

Here, `n_(1)=hat(i)+hat(j)+hat(k)andn_(2)=2hat(i)+3hat(j)+4hat(k)`
`and" "d_(1)=6andd_(2)=-5`
Hence, using relation `r*(n_(1)+lamdan_(2))=d_(1)+lamdad_(2)`, we get `r*[hat(i)+hat(j)+hat(k)+lamda(2hat(i)+3hat(j)+4hat(k))]=6-5lamda`
`rArrr*[(1+2lamda)hat(i)+(1+3lamda)hat(j)+(1+4lamda)hat(k)]=6-5lamda` . . . (i)
where, `lamda` is some real number.
Taking `r=xhat(i)+yhat(j)+zhat(k)`, we get
`xhat(i)+yhat(j)+zhat(k)`, we get
`(xhat(i)+yhat(j)+zhat(k))*[(1+2lamda)hat(i)+(1+3lamda)hat(j)+(1+4lamda)hat(k)]=6-5lamda`
`rArr" "(1+2lamda)x+(1+3lamda)y+(1+4lamda)z=6-5lamda`
`rArr" "(x+y+z-6)+lamda(2x+3y+4z+5)=0` . . . (ii)
Given that, the plane passes through the point (1,1,1), it must satisfy Eq. (ii) i.e. `(1+1+1-6)+lamda(2+3+4+5)=0`
`rArr" "lamda=(3)/(14)`
On putting the value of `lamda` in Eq. (i) , we get
`r*[(1+(3)/(7))hat(j)+(1+(9)/(14))hat(j)+(1+(6)/(7))hat(k)]=6-(15)/(14)`
`or" "r*((10)/(7)hat(i)+(23)/(14)hat(j)+(13)/(7)hat(k))=(69)/(14)`
`orr*(20hat(i)+23hat(j)+26hat(k))=69`