Question

Find the coordinates of the point where the line through (5, 1, 6) and

(3, 4, 1) crosses the YZ-plane

Answer 1

Answer:

point is ( 0  ,  17/2  , -13/2)

Explanation:

Equation of line Passing through

( 5 , 1 , 6)  &  ( 3 , 4 , 1)

= (x - 5) /(3-5)   = (y - 1)/(4 - 1)  = (z - 6)/(1 - 6)

= (5 - x)/2   = (y - 1)/3  = (6 - z)/5

(5 - x)/2   = (y - 1)/3  = (6 - z)/5  = k

=> 5 - x = 2k  

while x = 0

=> k = 5/2

y - 1  = 3k

=> y = 3k + 1  = 3(5/2) + 1 = 17/2

6-z = 5k

=> z = 6 - 5k  = 6 - 5(5/2)  = (12 - 25)/2 = -13/2

point is ( 0  ,  17/2  , -13/2)

Answer 2

Answer:

( $\mathbf{0}$, $\mathbf{\frac{17}{2}}$, $\mathbf{\frac{13}{2}}$)

Explanation:

In this question,

We need to find the coordinates of the point where the line passes through (5,1,6) and (3,1,4) and crosses the YZ-plane

Equation of Line Passing through two points

( 5 , 1 , 6)  &  ( 3 , 4 , 1)

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

= $\frac{(5 - x)}{2} = \frac{(y - 1)}{3} = \frac{(6 - z)}{5}$

$\frac{(5 - x)}{2} = \frac{(y - 1)}{3} = \frac{(6 - z)}{5} = k$         (say)

Therefore,

=> 5 - x = 2k  

Since, Line crosses the YZ-plane. Therefore,x = 0

=> k = $\frac{5}{2}$

Now,

y - 1  = 3k

=> y = 3k + 1  = 3($\frac{5}{2}$) + 1 = $\frac{17}{2}$

Similarly,

6-z = 5k

=> z = 6 - 5k  = 6 - 5($\frac{5}{2}$)  = $\frac{(12 - 25)}{2}\ =\ \frac{-13}{2}$

Therefore, Point is ( $\mathbf{0}$, $\mathbf{\frac{17}{2}}$, $\mathbf{\frac{13}{2}}$)