Question

The position vector of the point where the line $r=i-j+k+λ(i+j-k)$ meets the plane $r.(i+j+k)=5$ is –
A. $5i+j-k$
B. $5i+3j-3k$
C. $2i+j+2k$
D. $ 5i+j+k$

Answer 1

Hint: The vector equation of point will satisfy both equations, the equation of line $r=i-j+k+λ(i+j-k)$ as well as the equation of plane $r.(i+j+k)=5$ because it lies on both plane and line. Let the vector equation of the point is$ q=xi+yj+zk so q=r.$


Complete step by step solution: According to the question the vector equation of the plane is –
$ r.(i+j+k)=5 $ ----------(i)
the vector equation of the line is –
$ r=i-j+k+λ(i+j-k)$
$r=(1+λ)i-(1-λ)j+(1-λ)k $ ----------(ii)
Let the vector equation of point is –
$q=xi+yj+zk$ ----------(ii)
Since the point lies on both line and plane so it will satisfy both the equations thus –
$q=r$
$xi+yj+zk=(1+λ)i-(1-λ)j+(1-λ)k$
On comparing both the sides,
$x = (1 + λ), y = – (1 – λ) =(λ – 1)$, and$ z = (1 – λ)$
Since, $q=r$ so from equation (i)
$q.(i+j+k)=5
(xi+yj+zk).(i+j+k)=5
x(i.i)+y(j.j)+z(k.k)=5$
We know that, $(i.i)=(j.j)=(k.k)=1
x + y + z = 5$
Putting values of the $x$,$ y$, and $z$ in the equation
$(1 + λ) + (λ – 1) + (1 – λ) = 5
 λ + 1 = 5
λ = 5 – 1
λ = 4$
Putting value of$ λ$ in equation (ii)
$r=(1+4)i-(1-4)j+(1-4)k
r=5i-(-3)j+(-3)k
r=5i+3j-3k$
Since; we know that
$q=r
q=5i+3j-3k$
Hence, the position vector of the point is $q=5i+3j-3k$

Thus, Option (B) is correct.

Note: The position vector of a point is used to detect the position of that point with respect to origin in space. The direction of the position vector is always from origin towards the point. Position vector can be determined by the coordinates of the point.