Consider the points A(3, – 4, – 5) and 5(2, – 3,1) (i) Find the

1.	Consider the points A(3, – 4, – 5) and 5(2, – 3,1) (i) Find the vector and Cartesian equation of the Line passing through the points A and B. (ii) Find the point where the line crosses the XY Plane.
Answer» Let \(\bar{a}\) = 3i – 4j – 5k, b = 2i – 3j + k (i) Vector Equation is \(\bar{r}\) = \(\bar{a}\)+ λ(\(\bar{b}\) –\(\bar{a}\) )\(\bar r\) = 3i – 4j – 5k + λ( – i + j + 6k) Cartesian Equation is = \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6 }\) (ii) Let the point be (x, y, 0) \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6 }\) = \(x= \frac{13}{6},y = \frac{-19}{6 }\) Then the point on the XY Plane is \(\frac{13}{6},\frac{-19}{6}\),0

Consider the points A(3, – 4, – 5) and 5(2, – 3,1) (i) Find the vector and Cartesian equation of the Line passing through the points A and B. (ii) Find the point where the line crosses the XY Plane.

Answer»

Let \(\bar{a}\) = 3i – 4j – 5k, b = 2i – 3j + k

(i) Vector Equation is \(\bar{r}\) = \(\bar{a}\)+ λ(\(\bar{b}\) –\(\bar{a}\) )\(\bar r\)

= 3i – 4j – 5k + λ( – i + j + 6k)

Cartesian Equation is

= \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6 }\)

(ii) Let the point be (x, y, 0)

\(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6 }\)

= \(x= \frac{13}{6},y = \frac{-19}{6 }\)

Then the point on the XY Plane is

\(\frac{13}{6},\frac{-19}{6}\),0