Question
Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :
The correct answer is
$12$
The problem asks for the sum of all possible values of $r$ given a line $L$ and its distance from a point.
Line Properties Analysis
- The line $L$ passes through the point $P_1 = (-3, 5, 2)$.
- The line makes equal angles with the positive coordinate axes. This means its direction cosines $(l, m, n)$ satisfy $l=m=n$.
- From the direction cosine identity $l^2 + m^2 + n^2 = 1$, we get $3l^2 = 1$, which implies $l = \pm \frac{1}{\sqrt{3}}$.
- Therefore, the direction ratios $(a, b, c)$ of the line can be taken as $(1, 1, 1)$ or $(-1, -1, -1)$. We will use $\vec{v} = (1, 1, 1)$ as the direction vector. The magnitude is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Distance Calculation
The distance $d$ of a point $Q(x_0, y_0, z_0)$ from a line passing through $P_1(x_1, y_1, z_1)$ with direction vector $\vec{v}$ is given by the formula: $ d = \frac{|\vec{P_1Q} \times \vec{v}|}{|\vec{v}|} $
- The given point is $Q = (-2, r, 1)$.
- The vector $\vec{P_1Q}$ is calculated as $Q - P_1$: $ \vec{P_1Q} = (-2 - (-3), r - 5, 1 - 2) = (1, r-5, -1) $
- Calculate the cross product $\vec{P_1Q} \times \vec{v}$: $ \vec{P_1Q} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & r-5 & -1 \\ 1 & 1 & 1 \end{vmatrix} $ $ = \mathbf{i}((r-5)(1) - (-1)(1)) - \mathbf{j}(1(1) - (-1)(1)) + \mathbf{k}(1(1) - (r-5)(1)) $ $ = \mathbf{i}(r-5+1) - \mathbf{j}(1+1) + \mathbf{k}(1 - r + 5) $ $ = (r-4)\mathbf{i} - 2\mathbf{j} + (6-r)\mathbf{k} $
- Find the magnitude of the cross product: $ |\vec{P_1Q} \times \vec{v}| = \sqrt{(r-4)^2 + (-2)^2 + (6-r)^2} $ $ = \sqrt{(r^2 - 8r + 16) + 4 + (36 - 12r + r^2)} $ $ = \sqrt{2r^2 - 20r + 56} $
- The distance $d$ is: $ d = \frac{\sqrt{2r^2 - 20r + 56}}{\sqrt{3}} $
Solving for 'r'
We are given that the distance $d = \sqrt{\frac{14}{3}}$. Equating the two expressions for $d$: $ \frac{\sqrt{2r^2 - 20r + 56}}{\sqrt{3}} = \sqrt{\frac{14}{3}} $ Square both sides: $ \frac{2r^2 - 20r + 56}{3} = \frac{14}{3} $ Multiply by 3: $ 2r^2 - 20r + 56 = 14 $ Rearrange the terms to form a quadratic equation: $ 2r^2 - 20r + (56 - 14) = 0 $ $ 2r^2 - 20r + 42 = 0 $ Divide by 2: $ r^2 - 10r + 21 = 0 $ This is a quadratic equation in $r$.
Sum of Possible Values of 'r'
For a quadratic equation of the form $ar^2 + br + c = 0$, the sum of the roots (possible values of $r$) is given by $-\frac{b}{a}$.
- In the equation $r^2 - 10r + 21 = 0$, we have $a=1$, $b=-10$, and $c=21$.
- The sum of all possible values of $r$ is: $ \text{Sum} = - \frac{b}{a} = - \frac{-10}{1} = 10 $
Note: Using the direction vector $(-1, -1, -1)$ results in the same quadratic equation and hence the same sum of roots.
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