Find the vector and cartesian equations of the line passing

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Question 2:-  Find the vector and cartesian equations of the line passing through the point (2 , 1, 3) and perpendicular to the lines \dfrac{x - 1}{1} = \dfrac{y - 2}{2} = \dfrac{z - 3}{3} and \dfrac{x}{-3} = \dfrac{y}{2} = \dfrac{z}{5}.

Solution:- Let the cartesian equation of the line passing through (2, 1, 3) be

\dfrac{x - 2}{a} = \dfrac{y - 1}{b} = \dfrac{z - 3}{c}  .  .  . (i)

Since, line (i) iss perpendicular to given line

\dfrac{x - 1}{1} = \dfrac{y - 2}{2} = \dfrac{z - 3}{3}    .  .   .(ii)

and \dfrac{x}{-3} = \dfrac{y}{2} = \dfrac{z}{5}  .   .   . (iii)

∴ 1a + 2b + 3c = 0  .   .   . (iv)

-3a + 2b + 5c  = 0  .  .  . (v)

Solving the equation (iv) and (v) using partial fraction

\dfrac{a}{10 - 6} = \dfrac{b}{- 9 - 5} = \dfrac{c}{2 + 6}

⇒ \dfrac{a}{4} = \dfrac{b}{-14} = \dfrac{c}{8}

Substituting the value a, b and c in equation (i)

\dfrac{x - 2}{4} = \dfrac{y - 1}{-14} = \dfrac{z - 3}{8}

⇒ \dfrac{x - 2}{2} = \dfrac{y - 1}{-7} = \dfrac{z - 3}{4}

The vector form is

\vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 4\hat{k})

Other Three Dimensional Geometry Question

Question 1:-  Find the eqaaution of the line which intersect the lines \dfrac{x + 2}{1} = \dfrac{y - 3}{2} = \dfrac{z + 1}{4} and \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4} passes through the point (1, 1, 1)

Solution:- See ful solution

Question 3:- A line passing through the point A with position vector \vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k} is parallel to the vector \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}. Find the length of the perpendicular drawn on this line from a point P with position vector \vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}.                              [CBSE Panchkula 2015]

Solution:- See full solution

Case Study

 

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