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Question

Mathematics Question on Three Dimensional Geometry

Find the vector and the cartesian equations of the line that passes through the points(3,-2,-5),(3,-2,6).

Answer

Let the line passing through the points P (3,-2,-5) and Q(3,-2,6), be PQ.

Since PQ passes through P (3,-2,-5), its position vector is given by,

a→=3i^−2j^−5k^\overrightarrow a=3\widehat{i}-2\widehat j-5\widehat k
The direction ratios of PQ are given by, (3-3)=0, (-2+2)=0, (6+5)=11
The equation of the vector in the direction of PQ is

b→=0i^−0.j^+11k^=11k^\overrightarrow b=0 \widehat i-0. \widehat j+11 \widehat k=11 \widehat k

The equation of PQ in vector form is given by, r→=a→+λb→,λ∈R\overrightarrow r= \overrightarrow a+ \lambda \overrightarrow b, \lambda \in R
⇒r→=(3i^−2j^−5k^)+11λk^\Rightarrow \overrightarrow r=(3 \widehat i-2\widehat j-5\widehat k)+11 \lambda \widehat k

The equation of PQ in cartesian form is

x−x1a=y−y1b=z−z1ci.e.,x−30=y+20=z+511\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} i.e.,\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}