First slide

Planes in 3D

Question

consider a plane x+y-z=1 andpoint,4(1,2,-3).A line L has the equation x=1+3r,y=2-r and z=3+4r

Moderate

Question

The coordinate of a point B of line I 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)

Solution

The line $\frac{x - 1}{3} = \frac{y - 2}{- 1} = \frac{z - 3}{4} = r$
hence,
$\vec{A B} is parallel to x + y - z = 1$
$\begin{array}{l} \Rightarrow 3 r - r - 4 r - 6 = 0 or r = - 3 \\ B is (- 8, 5, - 9) \end{array}$

Question

The equation of the plane containing line L and point A has the equation

A

x-3y+5=0
B

x+3y-7=0
C

3x+y-5=0
D

3x-y-1=0

Solution

The equation of plane containing the line L is
$\begin{aligned} A (x - 1) + B (y - 2) + C (z - 3) = 0 \\ where 3 A - B + 4 C = 0 - - - - (i) \end{aligned}$
(i) also contains point A(1,2,-3).
Hence, C= 0and 3A=8.
the equation of plane $x - 1 + 3 (y - 2) = 0$ $or x + 3 y - 7 = 0$

Question

The distance between the points on the line which are at a distance of $4 / \sqrt{3}$ from the plane is

A

$4 \sqrt{26}$
B

20
C

$10 \sqrt{13}$
D

none of these

Solution

The distance of point (1 + 3r, 2-r,3 + 4r)from the plane is
$\frac{| 1 + 3 r + 2 - r - 3 - 4 r - 1 |}{\sqrt{1 + 1 + 1}} = \frac{| 2 r + 1 |}{\sqrt{3}} = \frac{4}{\sqrt{3}}$
$\Rightarrow r = \frac{3}{2}, - \frac{5}{2}$
Hence, the points are $A (\frac{11}{2}, \frac{1}{2}, \frac{10}{2})$ and $B (\frac{- 13}{2}, \frac{9}{2}, \frac{- 14}{2})$
$\Rightarrow A B = \sqrt{292}$

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