Test: Dimensional Geometry - 5 - Mathematics MCQ

Proof: We have

therefore the line joining (3, -1) and (2, 3) is perpendicular to the line joining (5, 2) and (9, 3).

Equations of Bisectors of the Angles between two non-parallel lines.
Let the straight lines
Ax+ By + C = 0 and
A'x + B'y + C = 0
be non-parallel. Then the bisectors are

Rules:
1. if C > 0, C' > 0 and AA' + BB' > 0, then (i) is the equation of obtuse - angle bisector (so the equation (ii) gives the acute - angle bisector).
2. If C > 0, C' > 0 and AA' * BB' < 0, then (i) is the equation of acute - angle bisector (so (ii) gives the obtuse - angle bisector)
3. If C and C' are of the same sign (cither both positive or both negative), then (i) is the bisector of that angle (acute or obtuse) in which the origin lies.
In Problem 42, the given equations are
3x-4y + 7= 0 and
12x - 5y = 8 = 0
Since C and C' are of opposite sign, therefore the bisector of the angle between (iii) and (iv), in which the origin lies in given by

The given straight lines are

Their point of intersection is obtained on solving equations (i) and (ii) and is

 The triangle is formed bv the straight lines

Their points of intersection are the vertices of the triangle and are (0, 0), (c, c) and (c, -c) The required area Δ is given by

A homogeneous equation of nth degree in x and y is a₀xⁿ + a₁x^n-1 + a₂x^n-2+ a₃x^n-3 .....+ a_n-1xy^n-1+ a_nyⁿ=0 and it represents at most n straight lines passing through the origin.

The lines are represented by

 The angle φ between the lines is given by

The condition that the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines is that af²+bg²+ch²+2jgh-abc=0

The required equation of lines is obtained bymaking equation

If A_x, A_y, A_z be the areas of projection of an area A on the three coordinate planes then A²=A_x² + A_y² + A_z² 
we have to find the area A of triangle ABC A_x =  the projection of A on yz plane 
A_y = the projection of A on zx plane 
A_z = the projection of A on xy plane 
∴ A²=A_x² + A_y² + A_z²

Let the coordinates of the vertices O, A, B, C o f tetrahedron be (α, β, γ), (a, 0, 0), (0, b, 0) and (0. 0, c) respectively. Then it is given that
OA is perpendicular to BC
and OB is perpendicular to CA
To prove that
OC is perpendicular to AB

The general equation of first degree in x, y, z namely

represents a plane, where at least one of A, B and C is non-zero.
Remark: 1. Let A ≠ 0. Then equation (i)

Therefore the general equation of a plane involves three independent (essential) arbitrary constant.
Hence if three independent conditions are given then the plane can be determined completely.
Remark : 2. If D = 0 then the plane will pass through the origin.

Let the coordinates of A, B and C be (a, 0, 0), (0, β, γ) and (0,0, y) respectively. Then the coordinates of the centroid of ΔABC is given by

In other words, the coordinates of A, B and C are
(3a, 0, O), (0, 3b, 0) and (0, 0, 3c) respectively.
Verify that the secoordinates of A, B and C satisfy equation in statement (c)

OP is perpendicular on the plane ABC

∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
Then
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.

Proof: OA is normal to the plane.

or ax + by + cz = a² + b² + c² (statement(al)) 
Further, Let P(x, y, z) be any point on the plane.
Then OA is perpendicular to PA.
=> (x-a)a + (y- b)b+(z- c)c=0
(This is statement (b))
.*. both (a) and (b) are correct.

Let P(x, y, z) be an arbitrary point on the plane. Then the line PA joining the points P(x, y, z) and A(2, -3, 1) lies on the plane and the d.r.’s of PA are
x - 2, y + 3, z- 1
Further, the direction ratios of the line joining points B(3, 4. - 1 ) and C(2, 1,5) are
3 - 2 , 4 - ( - 1 ),- 1 -5 
i.e. 1, 5, -6
Since BC is perpendicular to the plane and hence is perpendicular to line PA, therefore
( x - 2) 1 + (y + 3) 5 + ( z - 1) (- 6 ) = 0
or x + 5y - 6z + 19 = 0

The plane is parallel' to x-axis. Therefore, it is perpendicular to yzplane.
Hence its equation should be of the form 
By Cz + D = 0
Since, the plane passes through (2, 3, 1) and (4, -5, 3), therefore.