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:
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:
Show Hint
- 49
- 47
- 21
- 62
The Correct Option is B
Approach Solution - 1
2. Distance from the point $(1, 2, 3)$ to the line $L$: \[ \text{Distance} = 2\sqrt{17} \]
3. Use the distance formula: \[ \sqrt{(3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2} = 2\sqrt{17} \] \[ (3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2 = 68 \] \[ 17\lambda^2 - 17 = 0 \implies \lambda = \pm 1 \]
4. Determine the points $A$ and $B$: - For $\lambda = 1$: $A(9, 9, 5)$ - For $\lambda = -1$: $B(-3, -1, 9)$ 5. Calculate $\overrightarrow{OA} \cdot \overrightarrow{OB}$: \[ \overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(-1) + 5(9) = -27 - 9 + 45 = 47 \] Therefore, the correct answer is (2) 47.
Approach Solution -2
We have the line \(L: \dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}\). Let its parametric form be \(\vec{r}(t)=\vec{a}+t\vec{d}\), where \(\vec{a}=(6,7,7)\) and \(\vec{d}=(3,2,-2)\). A point \(P=(1,2,3)\) drops a perpendicular to \(L\) at foot \(F\). Points \(A,B\in L\) are each at distance \(2\sqrt{17}\) from \(F\). We need \(\overrightarrow{OA}\cdot \overrightarrow{OB}\).
Concept Used:
The foot \(F\) of the perpendicular from \(P\) to the line \(\vec{r}(t)=\vec{a}+t\vec{d}\) occurs at parameter
\[ t_0=\frac{\vec{d}\cdot(\vec{P}-\vec{a})}{\vec{d}\cdot \vec{d}}. \]
Moving a distance \(s\) along the line changes the parameter by \(\Delta t=\dfrac{s}{\|\vec{d}\|}\). If \(A,B\) are at distance \(2\sqrt{17}\) from \(F\), then \(|\Delta t|=\dfrac{2\sqrt{17}}{\|\vec{d}\|}\).
Step-by-Step Solution:
Step 1: Find the foot parameter \(t_0\).
\[ \vec{d}=(3,2,-2),\quad \vec{a}=(6,7,7),\quad \vec{P}-\vec{a}=(1-6,\,2-7,\,3-7)=(-5,-5,-4). \] \[ \vec{d}\cdot(\vec{P}-\vec{a})=3(-5)+2(-5)+(-2)(-4)=-15-10+8=-17,\quad \vec{d}\cdot\vec{d}=3^2+2^2+(-2)^2=9+4+4=17. \] \[ t_0=\frac{-17}{17}=-1. \]
Thus the foot \(F=\vec{r}(t_0)=\vec{a}+(-1)\vec{d}=(6,7,7)-(3,2,-2)=(3,5,9).\)
Step 2: Determine the parameters for \(A\) and \(B\).
\[ \|\vec{d}\|=\sqrt{17},\quad s=2\sqrt{17}\ \Rightarrow\ |\Delta t|=\frac{2\sqrt{17}}{\sqrt{17}}=2. \] \[ t_A=t_0+2=-1+2=1,\qquad t_B=t_0-2=-1-2=-3. \]
Step 3: Compute \(A=\vec{r}(1)\) and \(B=\vec{r}(-3)\).
\[ A=\vec{a}+1\cdot\vec{d}=(6,7,7)+(3,2,-2)=(9,9,5), \] \[ B=\vec{a}+(-3)\vec{d}=(6,7,7)+(-9,-6,6)=(-3,1,13). \]
Step 4: Compute the dot product \(\overrightarrow{OA}\cdot\overrightarrow{OB}=A\cdot B\).
\[ A\cdot B=(9,9,5)\cdot(-3,1,13)=9(-3)+9(1)+5(13)=-27+9+65=47. \]
Final Computation & Result
\(\overrightarrow{OA}\cdot \overrightarrow{OB}=47\).
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