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A line passing through the points $A(1,2,3)$ and $B(5,8,11)$ intersects the line $\vec{r}=4\hat{i}+\hat{j}+\lambda(5\hat{i}+2\hat{j}+\hat{k})$. Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.
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Q: A line passing through the points $A(1,2,3)$ and $B(5,8,11)$ intersects the line $\vec{r}=4\hat{i}+\hat{j}+\lambda(5\hat{i}+2\hat{j}+\hat{k})$. Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.

Question Analysis & Solution

Detailed Solution

Step 1: Find the equation of line AB

The line passes through $A(1,2,3)$ and $B(5,8,11)$. The direction vector is $\vec{v_1} = (5-1)\hat{i} + (8-2)\hat{j} + (11-3)\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}$. Simplifying, we use the direction ratio vector $\vec{d_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. The equation is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 4\hat{k})$.

Step 2: Find the point of intersection

Equating the two lines: $(1+2\mu, 2+3\mu, 3+4\mu) = (4+5\lambda, 1+2\lambda, \lambda)$. Solving the system: $3+4\mu = \lambda$. Substituting into the first two: $1+2\mu = 4+5(3+4\mu) \Rightarrow 1+2\mu = 19+20\mu \Rightarrow -18 = 18\mu \Rightarrow \mu = -1$. Thus, the point is $(1-2, 2-3, 3-4) = (-1, -1, -1)$.

Step 3: Find the direction of the perpendicular line

The direction of the required line is the cross product of the direction vectors of the two lines: $\vec{d_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{d_2} = 5\hat{i} + 2\hat{j} + \hat{k}$. $$\vec{d_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = \hat{i}(3-8) - \hat{j}(2-20) + \hat{k}(4-15) = -5\hat{i} + 18\hat{j} - 11\hat{k}$$

Step 4: Write the final equation

The line passes through $(-1, -1, -1)$ with direction vector $-5\hat{i} + 18\hat{j} - 11\hat{k}$. The equation is $\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + t(-5\hat{i} + 18\hat{j} - 11\hat{k})$.

Final Answer: Point: (-1, -1, -1); Equation: \vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + t(-5\hat{i} + 18\hat{j} - 11\hat{k})

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