Question

A student draws a plane and takes two points A and B as shown above. Answer the following questions.

I. What are the direction cosine of line joining A and B?



II. Write the equation of line joining A and B.


III. Direction ratios of the given plane are

a) 1, 1, 2
b) 1, 2, 1
c) 2, 1, 1
d) 1, 1, 1


IV. Point of intersection of the line joining the points A, B and the plane is

a) (-1, 1, 6)
b) (1, - 2, 7)
c) (3, -4, 5)
d) none of these


V. Distance between the points A and B is

a) 6 units
b) √18 units
c) √38 units
d) √37 units
​

Answer 1

Answer:

Step-by-step explanation:

Given:

• Point A (3, -4, -5)
• Point B (2, -3, 1)
• Equation of the plane = 2x + y + z = 7

Solution:

I.

The drs of the line joining A and B is given by,

$\sf drs\:of\: AB =<(2-3),(-3+4),(1+5)>$

$\sf drs\:of\:AB=<-1,1,6>$

Now the direction cosines (l, m, n) of a line is given by,

$\sf l=\pm \dfrac{a}{\sqrt{a^2+b^2+c^2} } , m=\pm\dfrac{b}{\sqrt{a^2+b^2+c^2} } ,n=\pm \dfrac{c}{\sqrt{a^2+b^2+c^2} }$

where a,b,c are the drs of the line.

Hence the dcs of the required line is given by,

$\sf l=\dfrac{-1}{\sqrt{1+1+36} } ,m=\dfrac{1}{\sqrt{1+1+36} } ,n=\dfrac{6}{\sqrt{1+1+36} }$

$\sf dcs \:of\: AB=<\dfrac{-1}{\sqrt{38} } ,\dfrac{1}{\sqrt{38} } ,\dfrac{6}{\sqrt{38} } >$

II.

Equation of a line joining two points is given by,

$\sf \dfrac{x-x_1}{x_2-x_1} =\dfrac{y-y_1}{y_2-y_1} =\dfrac{z-z_1}{z_2-z_1}$

Substitute the values,

$\sf \dfrac{x-3}{2-3} =\dfrac{y+4}{-3+4} =\dfrac{z+5}{1+5}$

$\sf \dfrac{x-3}{-1} =\dfrac{y+4}{1} =\dfrac{z+5}{6}$

This is the equation of the required line.

III.

The direction ratios of the normal to a given plane ax + by + cz = d are given by -  (a, b, c)

By given, the equation of the plane is 2x + y + z = 7

Hence the drs of normal to the plane are (2, 1, 1)

Therefore option c is correct.

IV.

The equation of the line joining the two points is given by,

$\sf \dfrac{x-3}{-1} =\dfrac{y+4}{1} =\dfrac{z+5}{6}$

Hence the coordinates of the general point which lies on the line is given by,

$\sf \dfrac{x-3}{-1} =\dfrac{y+4}{1} =\dfrac{z+5}{6}=\lambda$

x = -λ + 3

y = λ - 4

z = 6λ - 5

Since this point lies on the plane, it must satisfy the equation of the plane.

Therefore,

$\sf 2(-\lambda +3)+\lambda-4+6\lambda-5=7$

$\sf -2\lambda+6+\lambda-4+6\lambda-5=7$

$\sf 5\lambda-3=7$

$\sf 5\lambda=10$

$\sf \lambda=2$

Now put the value of λ in the above coordinates,

x = -2 + 3 = 1

y = 2 - 4 = -2

z = 12 - 5 = 7

Therefore the point of intersection is (1, -2, 7)

Therefore option b is correct.

V.

Distance between two points is given by,

$\sf Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Substitute the values,

$\sf Distance\:of\:AB=\sqrt{(2-3)^2+(-3+4)^2+(1+5)^2}$

$\implies \sf \sqrt{1+1+36}$

$\sf \implies \sqrt{38}\: units$

Therefore option c is correct.