Question

Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

Answer 1

Answer:

४

Step-by-step explanation:

Answer 2

Answer:

$\text{The required point is (1,-2,7)}$

Step-by-step explanation:

Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

$\text{The equation of the line through (3,-4,-5) and (2,-3,1) is }$

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$

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

$\text{Now, any point on this line can be written as }(-t+3,\:t-4,\:6t-5)$

$\text{since the line (1) crosses 2x+y+z-7=0, }$

$2(-t+3)+(t-4)+(6t-5)-7=0$

$-2t+6+(t-4)+(6t-5)-7=0$

$5t-10=0$

$\implies\:t=2$

$\therefore\:\text{The required point is (1,-2,7)}$