class 12 maths three dimensional geometry

Find the cartesian equation of the line which passes through the point $( - 2,4, - 5)$ and is parallel to the line given by $\cfrac{{x + 3}}{3} = \cfrac{{y - 4}}{5} = \cfrac{{z + 8}}{6}$.

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📘 Three Dimensional Geometry NCERT,EX.11.2,Q.6, Page .477 SA

Find the cartesian equation of the line which passes through the point $( - 2,4, - 5)$ and is parallel to the line given by $\cfrac{{x + 3}}{3} = \cfrac{{y - 4}}{5} = \cfrac{{z + 8}}{6}$.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

As the required line is parallel to the line

$\cfrac{{x + 3}}{3} = \cfrac{{y - 4}}{5} = \cfrac{{z + 8}}{6}$,

therefore , the line has direction ratios $< 3,5,6 >$.

Also, the line passes through $( - 2,4, - 5)$.

therefore , the equation of the line in (cartesian form) is

$\cfrac{{x - ( - 2)}}{3} = \cfrac{{y - 4}}{5} = \cfrac{{z - ( - 5)}}{6} \Rightarrow \cfrac{{x + 2}}{3}$

$= \cfrac{{y - 4}}{5} = \cfrac{{z + 5}}{6}$

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