A line drawn from the point A(1, 3, 2) parall | vedclass

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(A) The direction ratios of the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ are $2, 4, 1$. Since the required line is parallel to this line,its dir

Vedclass · Accepted Answer

$(\frac{1}{6}, \frac{4}{3}, \frac{19}{12})$

Vedclass · Answer

(A) The direction ratios of the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ are $2, 4, 1$. Since the required line is parallel to this line,its direction ratios are also $2, 4, 1$. The equation of the line passing through $A(1, 3, 2)$ with direction ratios $2, 4, 1$ is $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{1} = \lambda$. Any point $B$ on this line can be represented as $(2\lambda + 1, 4\lambda + 3, \lambda + 2)$. Since point $B$ lies on the plane $3x + y + 2z = 5$,we substitute the coordinates of $B$ into the plane equation: $3(2\lambda + 1) + (4\lambda + 3) + 2(\lambda + 2) = 5$. $6\lambda + 3 + 4\lambda + 3 + 2\lambda + 4 = 5$. $12\lambda + 10 = 5$. $12\lambda = -5 \Rightarrow \lambda = -\frac{5}{12}$. Substituting $\lambda = -\frac{5}{12}$ back into the coordinates of $B$: $x = 2(-\frac{5}{12}) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$. $y = 4(-\frac{5}{12}) + 3 = -\frac{5}{3} + 3 = \frac{4}{3}$. $z = -\frac{5}{12} + 2 = \frac{19}{12}$. Thus,the coordinates of point $B$ are $(\frac{1}{6}, \frac{4}{3}, \frac{19}{12})$.

$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are:

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