The coordinates of the point of intersection | vedclass

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Question

(A) Let the first line be $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{-1} = \lambda$. Any point on this line is $(3+\lambda, 5+2\lambda, 1-\lamb

Vedclass · Accepted Answer

$(2, 3, 2)$

Vedclass · Answer

(A) Let the first line be $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{-1} = \lambda$. Any point on this line is $(3+\lambda, 5+2\lambda, 1-\lambda)$.Let the second line be $\frac{x - 4}{2} = \frac{y - 2}{-1} = \frac{z - 4}{2} = \mu$. Any point on this line is $(4+2\mu, 2-\mu, 4+2\mu)$.For intersection, the coordinates must be equal:$3+\lambda = 4+2\mu \implies \lambda - 2\mu = 1$ $(i)$$5+2\lambda = 2-\mu \implies 2\lambda + \mu = -3$ (ii)Multiplying (ii) by $2$: $4\lambda + 2\mu = -6$ (iii)Adding $(i)$ and (iii): $5\lambda = -5 \implies \lambda = -1$.Substituting $\lambda = -1$ in $(i)$: $-1 - 2\mu = 1 \implies -2\mu = 2 \implies \mu = -1$.Check with $z$-coordinates: $1 - (-1) = 2$ and $4 + 2(-1) = 2$. Since $2 = 2$, the lines intersect.Point of intersection: $(3+(-1), 5+2(-1), 1-(-1)) = (2, 3, 2)$.

The coordinates of the point of intersection of the lines $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{-1}$ and $\frac{x - 4}{2} = \frac{y - 2}{-1} = \frac{z - 4}{2}$ are:

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