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A line L pass...

Question

A line L passing through origin is perpendicular to the lines
$L_{1} : \vec{r} = (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k} L_{2} : \vec{r} = (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}$
If the co-ordinates of the point in the first octant on $L_{2}$ at the distance of $\sqrt{17}$ from the point of intersection of $L & L_{1}$ are $(a, b, c)$ , then $18 (a + b + c)$ is equal to ______.

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Solution

Step1. Finding direction ration of the line ‘L’
Given equation of line,
$L_{1} : \vec{r} = (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k} L_{2} : \vec{r} = (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}$
$L_{1} : \frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 4}{2}$
Direction ratio of $L_{1} = 1, 2, 2$
$L_{2} : \frac{x - 3}{2} = \frac{y - 3}{2} = \frac{z - 2}{1}$
Direction ratio of $L_{2} = 2, 2, 1$
Therefore direction ratio of line ‘L’ is perpendicular to $L_{1} & L_{2}$
Direction ratio of ‘L’ is parallel to $L_{1} \times L_{2} = (- 2, 3, - 2)$
Therefore equation of line ‘L’
$L : \frac{x}{2} = \frac{y}{- 3} = \frac{z}{2}$
Step2 : Finding intersection of lines
Let,
$\Rightarrow L_{1} : \frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 4}{2} = μ$ &
$\Rightarrow L : \frac{x}{2} = \frac{y}{- 3} = \frac{z}{2} = λ$
Solving for $L & L_{1}$
$\Rightarrow λ = 1$
$\Rightarrow μ = - 1$
So intersection point be $P (2, - 3, 2)$
Step3. Finding the value of $18 (a + b + c)$
Let, $Q (2 ν + 3, 2 ν + 3, ν + 2)$ be required point on $L_{2}$
Now,
$\sqrt{{(2 v + 3 - 2)}^{2} + {(2 v + 3 + 3)}^{2} + {(v + 2 - 2)}^{2}} = P Q$
Also,
$\sqrt{{(2 v + 3 - 2)}^{2} + {(2 v + 3 + 3)}^{2} + {(v + 2 - 2)}^{2}} = \sqrt{17}$
Squaring on both side,
$\Rightarrow {(2 v + 3 - 2)}^{2} + {(2 v + 3 + 3)}^{2} + {(v + 2 - 2)}^{2} = 17 \Rightarrow 9 v^{2} + 28 v + 20 = 0 \Rightarrow v = - 2, - \frac{10}{9}$
$- 2$ (rejected)
Therefore
$\Rightarrow Q (3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}) \Rightarrow Q (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$
Now,
$\begin{array}{rcl} 18 (a + b + c) \\ = & 18 (\frac{7}{9} + \frac{7}{9} + \frac{8}{9}) \\ = & 44 \end{array}$

Hence, Answer is $18 (a + b + c) = 44$ .

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Similar questions

l

passing through the origin is perpendicular to the lines

l_{1} : (3 + t)^i + (- 1 + 2 t)^j + (4 + 2 t)^k, - \infty < t < \infty

l_{2} : (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)^k, - \infty < t < \infty

,
Then, the coordinate(s) of the point(s) on

l_{2}

at a distance of

\sqrt{17}

from the point of intersection of

l

l_{1}

L

passing through origin is perpendicular to the lines

L_{1} : \to r = (3 + t)^i + (- 1 + 2 t)^j + (4 + 2 t)^k

L_{2} : \to r = (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)^k

If the co-ordinates of the point in the first octant on

L_{2}

at the distance of

\sqrt{17}

from the point of intersection of

L

L_{1}

(a, b, c),

18 (a + b + c)

L

passing through origin is perpendicular to the lines

L_{1} : \to r = (3 + t)^i + (- 1 + 2 t)^j + (4 + 2 t)^k

L_{2} : \to r = (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)^k

If the co-ordinates of the point in the first octant on

L_{2}

at the distance of

\sqrt{17}

from the point of intersection of

L

L_{1}

(a, b, c),

18 (a + b + c)

A line l passing through the origin is perpendicular to the lines
$l_{1} : (3 + t)^i + (- 1 + 2 t)^j + (4 + 2 t)^k, - \infty < t < \infty l_{2} : (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)^k, - \infty < s < \infty$
Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of l and $l_{1}$ is (are)

l

passing through the origin is perpendicular to the lines

l_{1} : (3 + t)^i + (- 1 + 2 t)^j + (4 + 2 t)^k, - \infty < t < \infty

l_{2} : (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)^k, - \infty < s < \infty

Then, the coordinate(s) of the point(s) on

l_{2}

at a distance of

\sqrt{17}

from the point of intersection of

l

l_{1}

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