Find the image A‘ of the point A (2, 1, 2) in the line l

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Question 4:- Find the image A‘ of the point A (2, 1, 2) in the line l : \vec{r} = 4\hat{i} + 2\hat{j} + 2\hat{k} + \lamda(\hat{i} - \hat{j} - \hat{k}). Also, find the equation of the line joining AA’. Find the foot of perpendicular from point A on the line l.

Solution:- Given point A(2, 1, 2)

And line l : \vec{r} = 4\hat{i} + 2\hat{j} + 2\hat{k} + \lamda(\hat{i} - \hat{j} - \hat{k})

Cartesian equation of given line

\dfrac{x - 4}{1} = \dfrac{y - 2}{-1} = \dfrac{z - 2}{-1}

⇒ \dfrac{x - 4}{1} = \dfrac{y - 2}{-1} = \dfrac{z - 2}{-1} = \lamda(Let)

Let foot of perpendicular on line l P(λ + 4,  -λ + 2, -λ + 2)

Dr’s of the line PA (λ + 4 – 2, -λ + 2 – 1, -λ + 2 – 2)

= (λ + 2, -λ + 1, -λ)

Since line(l) is perpendicular to line PA then,

1(λ + 2) – 1(-λ + 1) – 1(-λ) = 0

⇒ λ + 2 + λ – 1 + λ = 0

⇒ 3λ + 1 = 0

⇒ λ = -1/3

Point P[-1/3 + 4,  -(-1/3) + 2, -(-1/3) + 2]

⇒ Foot of perpendicular on line is Point P[-11/3, 7/3, 7/3]

Let Point Q(a, b, c)

A(2, 1, 2) and foot of perpendicular P[-11/3, 7/3, 7/3]

x = \dfrac{x_1 + x_2 }{2}

⇒ \dfrac{-11}{3} = \dfrac{2 + a}{2}

⇒ \dfrac{-22}{3} = 2 + a

⇒ \dfrac{-22}{3} - 2 = a

⇒ \dfrac{-28}{3} = a

y = \dfrac{y_1 + y_2 }{2}

⇒ \dfrac{7}{3} = \dfrac{1 + b}{2}

⇒ \dfrac{14}{3} = 2 + b

⇒ \dfrac{14}{3} - 2 = b

⇒ \dfrac{8}{3} = b

z = \dfrac{z_1 + z_2 }{2}

⇒ \dfrac{7}{3} = \dfrac{2 + c}{2}

⇒ \dfrac{14}{3} = 2 + c

⇒ \dfrac{14}{3} - 2 = c

⇒ \dfrac{8}{3} = c

Image point A'[\dfrac{-28}{3}, \dfrac{8}{3}, \dfrac{8}{3}]

dr’s of the line AA’ (\dfrac{-28}{3} - 2, \dfrac{8}{3} - 1, \dfrac{8}{3} - 2)

dr’s of the line AA’ [\dfrac{-34}{3}, \dfrac{5}{3}, \dfrac{2}{3}]

Equation of lin AA’

\dfrac{x - 2}{-34/3} = \dfrac{y - 1}{5/3} = \dfrac{z - 2}{2/3}

⇒ \dfrac{x - 2}{-34} = \dfrac{y - 1}{5} = \dfrac{z - 2}{2}

Question 1:- Using integration, find the area of the region bounded by the line y = 5x + 2, the x- axis and the ordinates x = -2 and x = 2.                    [CBSE 2025]

Solution:- See full solution

Question 2:- Solve the following differential equation :

(1+x^2)\dfrac{dy}{dx} + 2xy = 4x^2.

Solution:- See full solution

Question 3:-  Solve the differential equation 2(y + 3) - xy\dfrac{dy}{dx} = 0; given y(1) = 2.

Solution:- See full solution

Question 5:- Find the shortest distance between the lines :

\dfrac{x + 1}{2} = \dfrac{y - 1}{1} = \dfrac{z - 9}{-3}

\dfrac{x - 3}{2} = \dfrac{y + 15}{-7} = \dfrac{z - 9}{5}.

Solution:- See full solution

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