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Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and is parallel to the vector $3\hat{i}+2\hat{j}-8\hat{k}$.
APPLY COMPETENCY 5 Marks
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MEDIUM
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Q: Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and is parallel to the vector $3\hat{i}+2\hat{j}-8\hat{k}$.

Question Analysis & Solution

Detailed Solution

Step 1: Find the point of intersection

Let the general points on the two lines be $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and $Q(5\mu+4, 2\mu+1, \mu)$. For intersection, $P=Q$: $$2\lambda+1 = 5\mu+4 \implies 2\lambda - 5\mu = 3$$ $$3\lambda+2 = 2\mu+1 \implies 3\lambda - 2\mu = -1$$ Solving these, multiply first by 3 and second by 2: $6\lambda - 15\mu = 9$ and $6\lambda - 4\mu = -2$. Subtracting gives $-11\mu = 11 \implies \mu = -1$. Substituting $\mu = -1$ into $3\lambda - 2(-1) = -1 \implies 3\lambda = -3 \implies \lambda = -1$. The point is $(-1, -1, -1)$.

Step 2: Formulate the Vector Equation

The line passes through $\vec{a} = -\hat{i} - \hat{j} - \hat{k}$ and is parallel to $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$. The vector equation is $\vec{r} = \vec{a} + \lambda\vec{b}$: $$\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$$

Step 3: Formulate the Cartesian Equation

Using the point $(x_1, y_1, z_1) = (-1, -1, -1)$ and direction ratios $(a, b, c) = (3, 2, -8)$, the equation is: $$\frac{x+1}{3} = \frac{y+1}{2} = \frac{z+1}{-8}$$

Final Answer: Vector: \vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}); Cartesian: \frac{x+1}{3} = \frac{y+1}{2} = \frac{z+1}{-8}

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