Question

L1;L2 are two lines. L1 has equation x = t + 1; y = t; z = t. L2 passes through (1; 1; 1), is orthogonal to L1, and is parallel to the plane x - y + z = 0

(a) Find the equation of L2. (Hint: L2 is parallel to the plane, so L2 is orthogonal to the normal vector of the plane. L2 is also orthogonal to L1).
(b) Find the distance between L1;L2.

Answer 1

Answer:

a) x = 1 -2t

y = 1

z = 1 + 2t

b)   $h=\frac{7\sqrt{26} }{26}$

Step-by-step explanation:

a) We need to find direction v of L

Let v = < a, b, c > ≠ 0

v ║ (Plane)

∴ v ⊥ (Normal(Plane))

∴ v ⊥ < 1, -1, 1 >

v ⊥ Direction(Line))

∴ v ⊥ < 1, 1, 1 >

∴ v ∝ < 1, -1, 1 > × < 1, 1, 1 >

Let v = < 1, -1, 1 > × < 1, 1, 1 > = < -2, 0, 2 >

Vector equation of L2 :

r(t) = < 1, 1, 1 > + t < -2, 0, 2 >

Parametric equation of L2 :

x = 1 -2t

y = 1

z = 1 + 2t

b) For L1:

A (1; 0; 0);     u = < 1, 1, 1 >

B (1; -1; 1);      v = < -2, 1, 2 >

AB = < 1-1, -1-0, 1-0 > = < 0, -1, 1 >

$V=det\left[\begin{array}{ccc}0&-1&1\\1&1&1\\-2&1&2\end{array}\right]=7$

Let  u × v = < 1, 1, 1 > × < -2, 1, 2 > = < 1, -4, 3 >

$Ab = \sqrt{(1)^{2} +(-4)^{2} +(3)^{2}} =\sqrt{26}$

Finally, the distance between L1 and L2 is

$h=\frac{V}{Ab}=\frac{7}{\sqrt{26}}=\frac{7\sqrt{26} }{26}$