Tag: point of intersection of a line and a plane

Quizzes (0) Questions (42)

Questions Related to point of intersection of a line and a plane

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

Find the point where the line of intersection of the planes $ x - 2y + z = 1$ and $x + 2y - 2z = 5$, intersects the plane $2x + 2y + z + 6 = 0$

  1. $(1, -2, -4)$
  2. $(0,0,-6)$
  3. $(1,0,-8)$
  4. $(-1,-1,-2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line of intersection of two planes is found by solving the system of equations. Testing the point (1, -2, -4) in the first plane: 1 - 2(-2) + (-4) = 1 + 4 - 4 = 1 (Correct). In the second plane: 1 + 2(-2) - 2(-4) = 1 - 4 + 8 = 5 (Correct). In the third plane: 2(1) + 2(-2) + (-4) + 6 = 2 - 4 - 4 + 6 = 0 (Correct).

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The line passing through the points $(5, 1,  a)$ and $(3, b, 1)$ crosses the $yz$-plane at the point $\left (0,\dfrac{17}{2},\dfrac{-13}{2}\right)$. Then,

  1. $a = 2, b = 8$
  2. $a = 4, b = 6$
  3. $a = 6, b = 4$
  4. $a = 8, b = 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of line passing through $(5, 1, a)$ and $(3, b, 1)$ is
$\dfrac{x-3}{5-3}= \dfrac{y-b}{1-b}= \dfrac{z-1}{a-1}$   ...(i)


Point $\left ( 0, \dfrac{17}{2}, -\dfrac{13}{2} \right )$ satisfies equation (i), we get

$-\dfrac{3}{2} = \dfrac{\dfrac{17}{2} -b}{1-b} = \dfrac{-\dfrac{13}{2}-1}{a-1}$

$\Rightarrow  a-1 = \dfrac{\left ( -\dfrac{15}{2} \right )}{\left ( -\dfrac{3}{2} \right )} = 5$
$\Rightarrow  a = 6$


Also,  $-3\left ( 1 - b \right )= 2 \left ( \dfrac{17}{2} - b\right )$

$\Rightarrow  3b - 3 = 17 - 2b$

$\Rightarrow  5b = 20   $

$  \Rightarrow  b = 4$