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.
Similar Questions
- Let a vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}, \lambda > 0$, make an obtuse angle with the vector $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle $\theta, \frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta) - \{\gamma\}$, then $\alpha + \beta + \gamma$ is equal to ________.
- Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $| \vec{CD} |^2$ is equal to :
Let $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.
Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :
- If the image of the point P($a$, 2, $a$) in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is P, then $a+b$ is equal to _____.
- If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a is equal to :
- Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}$. If the midpoint of the line segment PA is $(0, \frac{3}{4}, \frac{-1}{4})$, then the value of $a^2 + b^2 + \alpha^2$ is equal to :
- The square of the distance of the point ($-2$, $-8$, 6) from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2}$ is equal to:
- If $\left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}$, then the possible value(s) of $\alpha$ is (are)
- Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is :
- Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1$, $|\vec{b}|=4$, $|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^\circ$, then $|\vec{a} \cdot \vec{c}|$ is equal to
Important Questions from Vectors and 3D Geometry
- Let a vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}, \lambda > 0$, make an obtuse angle with the vector $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle $\theta, \frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta) - \{\gamma\}$, then $\alpha + \beta + \gamma$ is equal to ________.
- Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $| \vec{CD} |^2$ is equal to :
Let $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.
Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :
- If the image of the point P($a$, 2, $a$) in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is P, then $a+b$ is equal to _____.
- If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a is equal to :
