The position vector of the point of intersect | vedclass

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(C) Let the two lines be $L_1$ and $L_2$. $L_1$ passes through $A(2, -1, 6)$ and $B(3, -1, -7)$. The direction vector is $\vec{v_1} = (3-2)\hat{i} + (

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$3 \hat{i}-\hat{j}-7 \hat{k}$

Vedclass · Answer

(C) Let the two lines be $L_1$ and $L_2$. $L_1$ passes through $A(2, -1, 6)$ and $B(3, -1, -7)$. The direction vector is $\vec{v_1} = (3-2)\hat{i} + (-1 - (-1))\hat{j} + (-7-6)\hat{k} = \hat{i} - 13\hat{k}$. The equation of $L_1$ is $\vec{r} = (2\hat{i}-\hat{j}+6\hat{k}) + \lambda(\hat{i} - 13\hat{k})$. $L_2$ passes through $C(2, 1, -6)$ and $B(3, -1, -7)$. The direction vector is $\vec{v_2} = (3-2)\hat{i} + (-1-1)\hat{j} + (-7 - (-6))\hat{k} = \hat{i} - 2\hat{j} - \hat{k}$. The equation of $L_2$ is $\vec{r} = (2\hat{i}+\hat{j}-6\hat{k}) + \mu(\hat{i} - 2\hat{j} - \hat{k})$. Since both lines pass through the point $B(3, -1, -7)$, this point is the intersection point. Thus, the position vector is $3\hat{i} - \hat{j} - 7\hat{k}$.

The position vector of the point of intersection of the line joining the points $2 \hat{i}-\hat{j}+6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$, and the line joining the points $2 \hat{i}+\hat{j}-6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$ is:

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