3-D Geometry: Planes in Space(15 Dec) - JEE MCQ

Internal bisector of an angle divides the opposite side in the ratio of adjacent sides

Coordinate of are
Length

Here, intercepts made with the coordinate axes X,Y and Z are 1/3 , 1/4 and 1/5 respectively. Intercept form of a plane is

Let P(α, β, γ) be the foot of the perpendicular from the origin O(0,0,0) to the plane So, the plane passes through P(α,β,γ) and is perpendicular to OP. Clearly direction ratios of OP i.e., normal to the plane are α,β,γ. Therefore, equation of the plane is α(x − α) + β(y − β) + γ(z − γ) = 0
This plane passes through the fixed point (1, 2, 3), so α(1 − α) + β(2 − β) + γ(3 − γ) = 0
or α² + β² + γ² − α − 2β − 3γ = 0
Generalizing α,β and γ, locus of P(α,β,γ) is x² + y2 + z² − x − 2y − 3z = 0

Let be the image of the point in the plane , then is normal to the plane ∴ direction ratios of are Since passes through and has direction ratios
∴ Equation of is

Let {l, m,n} be the direction -cosines of PQ, then 3l − m + n = 0 and 5l + m + 3n = 0
i.e
Now a plane ⊥ to PQ will have l,m,n as the coefficients of x,y and z
Hence the plane ⊥ to PQ is x + y − 2z = λ It passes through (2,1,4); ∴2 + 1−2.4 = λ i.e λ = −5 Hence the required plane is x + y−2z = −5

A(0,b,c) in yz -plane and B(a,0,c) in zx -plane. Plane through O is px + qy + rz = 0. It passes through A and B.
∴ 0p + qb + rc = 0 and pa + 0q + rc = 0

⇒ p = bck, q = cak and r = −abk.
Hence required plane is bcx + cay − abz = 0.

The planes bx − ay = n, cy − bz = l and az − cx = m intersect in a line, if al + bm + cn = 0.

Let be the fixed point on the variable plane

NowD. R 's of OM are x − 0, y − 0, z − 0 i.ex, y,z
D.R.'s of MA are x − a, y − b, z − c.
Since OM perpendicular MA
x(x − a)+y(y − b)+z(z − c) = 0
⇒ x² + y² + z² − ax − by − cx = 0

Given planes are

and

Since both planes are parallel and
and
Distance
Distance

Let equation of plane through z-axis is

It is given that this plane is parallel to the line

Since the plane parallel to the line




which is required equation of plane.