Test: 3-D Geometry-Equation of Lines (14 Dec) - JEE MCQ

In the new frame x′ = x − x₁, y′ = y − y₁, z′ = z − z₁, where (x₁, y₁,z₁) is shifted origin.
⇒ x′ = 0 − 1 = −1, y′ = 4 − 2 = 2 = 2, z′ = 5 + 3 = 8
Hence, the coordinates of the point with respect to the new coordinates frame are (−1, 2, 8)

We have, for the point where the line intersects the curve.
Therefore,
and
and
Put these value in , we get,

Given equation of line is

Let it is foot of perpendicualr So, d.r.'s of line is

D.r.'s of given line is and both lines are

∴ Point is (−4,1,−3).[ Substituting λ=1 in (i) ]

Equation of line is Any point on this line is (K, 2K + 1, 3K + 2). If this is the foot of perpendicular from then of this perpendicular are
Now, using Condition of perpendicularity we have

Hence, Required foot of perpendicular is

Let the line be 
If line (i) intersects with the line 

Since line is parallel to the plane, vector
is perpendicular to the normal to
the plane
⇒  or

Equation of plane passing through intersecting the planes  and
Since it is passing through origin, so .
Hence the required equation is