Question
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and
parallel to the line $\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}$
parallel to the line $\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}$
ratios 3, 5, 6 as it is parallel to the given line. It passes through the point (-2, 4, -5). The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{2-z_1}{c}$
$\therefore$ the required cartesian equations of the line are
$\frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6}$
i.e. $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\cos \left(60^{\circ} 30^{\prime}\right)$, given that $1^{\circ}=0.0175^{\circ}, \sqrt{ } 3=1.732$
$\int \sqrt{\frac{10+x}{10-x}} \cdot d x$
$\sin \theta=-\frac{1}{2}$