Hence, the centroid

λ = cos $\alpha$ , m = cos $\beta$ and n = cos $\gamma$ are called direction cosines of line L.

x₂ – x₁, y₂ – y₁, z₂ – z₁ or x₁ – x₂, y₁ – y₂, z₁ – z₂

If the planes A₁x+ B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0 are parallel, then

P( x₁, y₁, z₁) and Q( x₂, y₂, z₂) are

where

Consider the two lines:

is given by

Let

Then the Cartesian equation of a plane is,

The direction ratios of the normal to the planes are A₁, B₁, C₁ and A₂, B₂, C₂.

< 1,1,2 > and <-3, -4,1 >.

and

and check whether the lines are parallel or perpendicular. (C.B.S.E. 2011)

units from the origin and its normal vector from the origin is 2$\hat{i}$−3$\hat{j}$+4$\hat{k}$. Also, find its cartesian form. (N.C.E.R.T.)

$\overrightarrow{r}$=($\hat{i}$−$\hat{j}$+$\hat{k}$)+λ(2$\hat{i}$−$\hat{j}$+3$\hat{k}$) and the plane $\overrightarrow{r}$⋅(2$\hat{i}$+$\hat{j}$−$\hat{k}$)=4 Also, find whether the line is parallel to the plane or not .

$\frac{1-x}{3}$=$\frac{7y-14}{\lambda }$=$\frac{z-3}{2}$ and $\frac{7-7x}{3\lambda }$=$\frac{y-5}{1}$=$\frac{6-z}{5}$ are at right angles. Also, find whether the lines are intersecting or not

cos² α + cos² β + cos² γ + cos² δ= $\frac{4}{3}$. (N.C.E.R.T.)

$\frac{x+1}{2}$=$\frac{1-y}{4}$=$\frac{z+2}{1}$

Hence, find the shortest distance between the lines. (C.B.S.E. Sample Paper 2018-19)

l² + m² + n² = 1

n² = 1 – $\frac{1}{2}$

n² = $\frac{1}{2}$

n = $\frac{1}{\sqrt{2}}$

Thus, cos α = $\frac{1}{\sqrt{2}}$

Hence, α = 45° or $\frac{\pi }{4}$

Its direction-ratios are <2,-1,2>.

Hence, its direction- cosine are:

Since α, β, γ are direction-angles of a line,

∴ cos2 α + cos2 β + cos2γ = 1

⇒ 1 + cos2α + 1 + cos2β + 1 + cos2γ = 2

⇒ cos 2α + cos 2β + cos 2γ + 1 = 0, which is true.

The given plane is2x + y – z = 5

Its intercepts are $\frac{x}{5/2}$, 5 and -5.

Hence, the length of the intercept on the x-axis is $\frac{x}{5/2}$

Solution:

Length of the perpendicular from P(3, -4,5) on the z-axis

Direction cosines of the line are:

< cos 90°, cos 135°, cos 45° >

The equations of the line through A (3,4,1) and B (5,1,6) are:

Any point on (1) is (3 + 2k,4- 3k, 1 + 5k) …. (2)

This lies on xy-plane (z = 0).

∴ 1 + 5k = 0 ⇒ k = −$\frac{1}{5}$

Putting in (2), [ 3-$\frac{2}{5}$, 4 +$\frac{3}{5}$, 1-1)

i.e. ($\frac{13}{5}$, $\frac{23}{5}$, 0)

which are the reqd. co-ordinates of the point.

The vector equation of the line is $\overrightarrow{r}$ =$\overrightarrow{a}$+λ$\overrightarrow{m}$

i.e., $\overrightarrow{r}$ =(3$\hat{i}$+4$\hat{j}$+5$\hat{k}$)+λ(2$\hat{i}$+2$\hat{j}$−3$\hat{k}$)

The given lines can be rewritten as:

Here < 2,7, – 3 > and < -1,2,4 > are direction- ratios of lines (1) and (2) respectively.

Hence, the given lines are perpendicular.

Vector equation of the line passing through

∴ Line (1) is perpendicular to z-axis.

Dividing (1) by 7,

Vector in the direction of first line

Vector in the direction of second line

∴θ, the angle between two given lines is given by:

The given line is:

If ‘θ’ is the angle between the line and the plane,

Then

Hence, the line is parallel to the plane.

(i) The given lines are

These are perpendicular if:

Hence λ = 1.

(ii) The direction cosines ofline(1) are <-3,1,2>

The direction cosines of line (2) are < -3,1, -5 >

Clearly, the lines are intersecting.

Comparing given equations with:

Let O be the origin and OA, OB, OC (each = a) be the axes.

Thus the co-ordinates of the points are :

O (0,0,0), A (a, 0,0), B (0, a, 0), C (0,0, a),

P (a, a, a), L (0, a, a), M (a, 0, a), N (a, a, 0).

Here OP, AL, BM and CN are four diagonals.

Let < l, m, n > be the direction-cosines of the given line.

Now direction-ratios of OP are:

<a-0,a-0,a-0>i.e.<a,a,a>

i.e. < 1,1,1 >,

direction-ratios of AL are:

<0-a, a-0, a-0> i.e. <-a,a,a>

i.e. <-l, 1,1 >,

direction-ratios of BM are:

<a-0,0-a, a-0>

i.e. <a,-a,a> i.e. < 1,-1, 1 >

and direction-ratios of CN are:

<a-0,a-0,0-a> i.e. <a,a,-a>

i.e. < 1,1,-1 >.

Thus the direction-cosines of OP are:

the direction-cosines of AL are:

the direction-cosines of BM are:

and the direction-cosines of CN are:

If the given line makes an angle ‘a’ with OP, then:

Squaring and adding (1), (2), (3) and (4), we get:

cos² α + cos² β + cos² γ + cos²δ

+ (l-m + n)² + (l + m — n)²]

The two given lines are:

Let <a, b, c> be the direction-ratios of the normal to the plane containing line (1).

∴ Equation of the plane is:

a(x- l) + b(y-4) + c(z-4) …(3),

where 3a + 2b – 2c = 0 …(4)

[∵ Reqd. plane contains line (1)] and 2a – 4b + 1.c = 0

[∵ line (1) a parallel to the reqd. plane] Solving (4) and (5),

Putting in (3),

6k(x- 1) + 7k(y – 4) + 16k(z – 4) = 0

= 6(x – 1) + 7(y – 4) + 16(z – 4) =0

[∵k ≠ 0]

⇒ 6x + 7y+ 16z-98 = 0,

which is the required equation of the plane.

Now, S.D. between two lines = perpendicular distance of (-1,1, – 2) from the plane

6(–1) + 7(1) +16(-2) – 98

V(6)²+(7)²+(16)²

-6 + 7-32-98 V³⁶ + 49 + 256

(i) Cartesian equations

Any plane through (2,2, -1) is:

a(x – 2) + b(y- 2) + c(z + 1) = 0 … (1)

Since the plane passes through the points (3,4,2) and (7,0,6),

∴ a(3 – 2) + b(4 – 2) + c(2 +1) = 0

and a(7 – 2) + b(0 – 2) + c(6 + 1) = 0

⇒ a + 2b + 3c = 0 …(2)

and 5a – 2b + 7c = 0 …(3)

∴ a = 5k,b = 2k and c = -3k,

Putting the values of a, b, c in (1), we get:

5k(x – 2) + 2k(y – 2) – 3k(z + 1) = 0

⇒ 5(x-2) + 2(y-2)-3(z+ 1) =0[∵ k ≠ 0]

=» 5x- 10 + 2y-4-3z-3 = 0

=» 5x + 2y-3z-17 = 0, …(4)

which is the reqd. Cartesian equation.

(ii) Any plane parallel to (4) is

5x + 2y – 3z + λ – 0 … (5)

Since it passes through (4, 3,1),

5(4) + 2(3) – 3(1) + λ = 0

⇒ 20 + 6 — 3 + λ = 0

⇒λ = -23.

Putting in (5), 5x + 2y – 3z – 23 = 0, which is the reqd. equation.