Question

Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:

• A. 49
• B. 47
• C. 21
• D. 62

Hint:

Use the distance formula to find the points on the line.

Answer 1

The Correct Option is B

1. Define points $A$ and $B$ with parameters $\lambda$ and $\mu$ respectively: - $A(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$ - $B(3\mu + 6, 2\mu + 7, -2\mu + 7)$
2. The distance from point $(1, 2, 3)$ to line $L$ is given as $2\sqrt{17}$.
3. Applying the distance formula leads to the equation: \[ \sqrt{(3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2} = 2\sqrt{17} \] Squaring both sides gives: \[ (3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2 = 68 \] Simplifying the equation results in: \[ 17\lambda^2 - 17 = 0 \implies \lambda = \pm 1 \]
4. Determine the coordinates of points $A$ and $B$ using the calculated values of $\lambda$: - For $\lambda = 1$, point $A$ is $(9, 9, 5)$. - For $\lambda = -1$, point $B$ is $(-3, -1, 9)$. 5. Calculate the dot product of vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$: \[ \overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(-1) + 5(9) = -27 - 9 + 45 = 47 \] The final result is 47.