Question

The co-ordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

• A. 10, -1, 15
• B. -5, 4, -5
• C. 4, 1, 7
• D. -8, 5, -9

Answer 1

Answer: (D)

(-8, 5, -9)

Answer 2

Let the point B on line L have coordinates $(1+3r, 2-r, 3+4r)$ for some parameter $r$.
The coordinates of point A are $(1, 2, -3)$.
The vector $\vec{AB}$ is given by $B - A = (1+3r - 1, 2-r - 2, 3+4r - (-3)) = (3r, -r, 6+4r)$.
The equation of the given plane is $x+y-z=1$. The normal vector to this plane is $\vec{n} = (1, 1, -1)$.
For the line segment AB to be parallel to the plane, the vector $\vec{AB}$ must be perpendicular to the normal vector $\vec{n}$. Their dot product must be zero.
$\vec{AB} \cdot \vec{n} = 0$
$(3r)(1) + (-r)(1) + (6+4r)(-1) = 0$
$3r - r - 6 - 4r = 0$
$-2r - 6 = 0$
$-2r = 6$
$r = -3$
Substitute the value of $r = -3$ into the parametric equations of line L to find the coordinates of point B:
$x = 1 + 3(-3) = 1 - 9 = -8$
$y = 2 - (-3) = 2 + 3 = 5$
$z = 3 + 4(-3) = 3 - 12 = -9$
The coordinates of point B are $(-8, 5, -9)$.