Test: Dimensional Geometry - 3 - Mathematics MCQ

Point in polar coordiantes

Points in Cartesian coordinates

Clearly the straighi line which makes an angle of 90°' with the positive direction of the x-axis and passes through the origin is ij-nxis and hence its equations given by x = 0

Note that the equation of a straight line which makes an angle 0 with the positive direction of the x-axis and cuts an intercept equal to c on y-axis, is given by

y= mx + c,
where m = tan θ.
In this problem,
c = 5
and
θ= 135° ⇒ w = tan 135° = -1 Required equation is given by
y = -x+5 
or y + x = 5

and so on.

Figure shows the straight line such that
(i) P is the perpendicular from the origin on the straight line.

(ii) Perpendicular makes an angle α with positive direction of the axis of x.
Its equation is given by x cos α + y sin α = p ...(i)
The required intercept on y-axis is obtained by substituting x = 0 in equation (i).
⇒ y sin a = p
=> y = p cosecor.

The general equation of a straight line is given by
Ax + By = C,
where A. B and C are constants.
Remark: Note that the equation

does represents a straight line. But it is not correct to say that only the equations of this type will represent a straight line. This is the intercept form of a straight line.

Proof: The equation of a st raight line having equal intercepts on the coordinate axis is given by

where a is the intercept on x or y-axis.
Since the. straight line (i) passes through the point (2, 3), therefore the coordinates of this point will satisfy equation (i).

If p is the length of the perpendicular from the origin on the line
5x + 12y - 26 = 0
then
​

let the given line joins the points P(x₁ ,y₁, z₁ ) and Q(x₂ ,y₂, z₂)
Then since the d.c.'s of x-axis are [1, 0,0]. therefore the projection of PQ on x-axis = 1 (x₂ - x₁ )
=> x₂ - x₁ =12
Similarly y₂ - y₁ = 4
and z₂ - z₁= 3
Required length r of the line PQ is given by 

D.C.'s of the line joining the points (4, 3, -5) and (-2, 1,-8) are


d.r.’s are 6,2,3

Direction cosines of the line joining A{3, -4, 7) and B(0, 2, 5) are given by


Now the required projection of line joining P(6, 3, 2) and Q(5, 1. 4) on the line AB is given by

Let the lines OP and OQ be parallel to the given lines, so that the direction cosines of OP are [l₁m₁n₁] and the direction cosines of OQ are [l₂m₂n₂]
Clearly




Remark-3: Condition of two lines being perpendicular.
If the two lines are perpendicular, then cosθ = l₁l₂+m₁m₂+n₁n₂= o
Remark-4: Condition of two lines being parallel. If the two lines are parallel, then 

Proof- Let (l,m,n) be the direction cosines of a line which is normal to the plane containing the lines with d.c.'s (l₁,m₁,n₁) and (l₂,m₂,n₂) since a plane can always be drawn containing any two concurrent lines] 
=> l₁l + m₁m + n₁n = 0 ...(i)
and l₂l + m₂m + n₂n = 0 ...(ii)

If the line with d.c.'s (l₃,m₃,n₃) also lies on this plane, then
l₃l + m₃m + n₃n = 0 ...(iii)
For a non-trivial solution of equations (i), (ii) and (iii) for l,m,n we get

Let the d.c.’s of the line which is perpendicular to the two lines with d.r.’s (1,-2, -2) and (0, 2, 1) be [l,m,n]
Now the d.c.'s of the line with d.r.'s (1,-2,-2)

Let[l,m,n] be the d.c.'s of the line which is perpendicular to the two line having d,c.'s as

Let D(a,b,c) be foot of perpendicular from A(1, 8, 4) on the line joining B(0, -11, 4) and C(2, -3. 1). Then the direction ratio of DC are proportional to the d.r.'s of BC.


∴ The coordinates of D are obtained by substituting k=1 in (1)
∴ Coordinates of D are (4,5,-2)