The question
(a) An insect is crawling along the line r ˉ = 6 i ^ + 2 j ^ + 2 k ^ + λ ( i ^ − 2 j ^ + 2 k ^ ) \bar r = 6\hat i + 2\hat j + 2\hat k + \lambda(\hat i - 2\hat j + 2\hat k) r ˉ = 6 i ^ + 2 j ^ + 2 k ^ + λ ( i ^ − 2 j ^ + 2 k ^ ) r ˉ = − 4 i ^ − k ^ + μ ( 3 i ^ − 2 j ^ − 2 k ^ ) \bar r = -4\hat i - \hat k + \mu(3\hat i - 2\hat j - 2\hat k) r ˉ = − 4 i ^ − k ^ + μ ( 3 i ^ − 2 j ^ − 2 k ^ )
OR
(b) The equations of motion of a rocket are:
x = 2 t , y = − 4 t , z = 4 t x = 2t, y = -4t, z = 4t x = 2 t , y = − 4 t , z = 4 t
r ⃗ = 20 i ^ − 10 j ^ + 40 k ^ + μ ( 10 i ^ − 20 j ^ + 10 k ^ ) \begin{aligned} \vec r &= 20\hat i - 10\hat j + 40\hat k \\ &\quad + \mu(10\hat i - 20\hat j + 10\hat k) \end{aligned} r = 20 i ^ − 10 j ^ + 40 k ^ + μ ( 10 i ^ − 20 j ^ + 10 k ^ ) Exam answer Show all explanations
(a) The direction ratios 1 , − 2 , 2 1, -2, 2 1 , − 2 , 2 3 , − 2 , − 2 3, -2, -2 3 , − 2 , − 2
Why?
Proportional direction ratios would mean parallel lines. Here 1 3 ≠ − 2 − 2 \frac13 \ne \frac{-2}{-2} 3 1 = − 2 − 2
For two lines that are not parallel, the shortest distance is measured along the common perpendicular: the segment PQ that is perpendicular to both lines. So we find P and Q, then PQ.
The position vector of P on the first line is
( 6 + λ ) i ^ + ( 2 − 2 λ ) j ^ + ( 2 + 2 λ ) k ^ for some λ \begin{aligned} &(6 + \lambda)\hat i + (2 - 2\lambda)\hat j \\ &\quad + (2 + 2\lambda)\hat k \text{ for some } \lambda \end{aligned} ( 6 + λ ) i ^ + ( 2 − 2 λ ) j ^ + ( 2 + 2 λ ) k ^ for some λ Why?
Every point of the line r ⃗ = a ⃗ + λ b ⃗ \vec r = \vec a + \lambda\vec b r = a + λ b a ⃗ + λ b ⃗ \vec a + \lambda\vec b a + λ b λ \lambda λ
i ^ : 6 + λ ( 1 ) j ^ : 2 + λ ( − 2 ) k ^ : 2 + λ ( 2 ) \begin{aligned} \hat i&:\ 6 + \lambda(1) \\ \hat j&:\ 2 + \lambda(-2) \\ \hat k&:\ 2 + \lambda(2) \end{aligned} i ^ j ^ k ^ : 6 + λ ( 1 ) : 2 + λ ( − 2 ) : 2 + λ ( 2 ) Q is written in the same way, with μ \mu μ
The position vector of Q on the second line is
( − 4 + 3 μ ) i ^ + ( − 2 μ ) j ^ + ( − 1 − 2 μ ) k ^ for some μ \begin{aligned} &(-4 + 3\mu)\hat i + (-2\mu)\hat j \\ &\quad + (-1 - 2\mu)\hat k \text{ for some } \mu \end{aligned} ( − 4 + 3 μ ) i ^ + ( − 2 μ ) j ^ + ( − 1 − 2 μ ) k ^ for some μ P Q → = ( − 10 + 3 μ − λ ) i ^ + ( − 2 μ − 2 + 2 λ ) j ^ + ( − 3 − 2 μ − 2 λ ) k ^ \begin{aligned} \overrightarrow{PQ} &= (-10 + 3\mu - \lambda)\hat i \\ &\quad + (-2\mu - 2 + 2\lambda)\hat j \\ &\quad + (-3 - 2\mu - 2\lambda)\hat k \end{aligned} PQ = ( − 10 + 3 μ − λ ) i ^ + ( − 2 μ − 2 + 2 λ ) j ^ + ( − 3 − 2 μ − 2 λ ) k ^ Why?
P Q → \overrightarrow{PQ} PQ
( − 4 + 3 μ ) − ( 6 + λ ) ( − 2 μ ) − ( 2 − 2 λ ) ( − 1 − 2 μ ) − ( 2 + 2 λ ) \begin{aligned} &(-4 + 3\mu) - (6 + \lambda) \\ &(-2\mu) - (2 - 2\lambda) \\ &(-1 - 2\mu) - (2 + 2\lambda) \end{aligned} ( − 4 + 3 μ ) − ( 6 + λ ) ( − 2 μ ) − ( 2 − 2 λ ) ( − 1 − 2 μ ) − ( 2 + 2 λ ) Open each bracket. The minus sign changes every sign inside the second bracket:
i ^ : − 4 + 3 μ − 6 − λ = − 10 + 3 μ − λ j ^ : − 2 μ − 2 + 2 λ k ^ : − 1 − 2 μ − 2 − 2 λ = − 3 − 2 μ − 2 λ \begin{aligned} \hat i&:\ -4 + 3\mu - 6 - \lambda \\ &= -10 + 3\mu - \lambda \\[4pt] \hat j&:\ -2\mu - 2 + 2\lambda \\[4pt] \hat k&:\ -1 - 2\mu - 2 - 2\lambda \\ &= -3 - 2\mu - 2\lambda \end{aligned} i ^ j ^ k ^ : − 4 + 3 μ − 6 − λ = − 10 + 3 μ − λ : − 2 μ − 2 + 2 λ : − 1 − 2 μ − 2 − 2 λ = − 3 − 2 μ − 2 λ Since PQ is perpendicular to both the lines,
( − 10 + 3 μ − λ ) + ( − 2 μ − 2 + 2 λ ) ( − 2 ) + ( − 3 − 2 μ − 2 λ ) 2 = 0 ⇒ 3 μ − 9 λ − 12 = 0 ⇒ μ − 3 λ = 4 … ( i ) \begin{aligned} &(-10 + 3\mu - \lambda) \\ &\quad + (-2\mu - 2 + 2\lambda)(-2) \\ &\quad + (-3 - 2\mu - 2\lambda)2 = 0 \\ \Rightarrow\ &3\mu - 9\lambda - 12 = 0 \\ \Rightarrow\ &\mu - 3\lambda = 4 \quad\ldots(\text{i}) \end{aligned} ⇒ ⇒ ( − 10 + 3 μ − λ ) + ( − 2 μ − 2 + 2 λ ) ( − 2 ) + ( − 3 − 2 μ − 2 λ ) 2 = 0 3 μ − 9 λ − 12 = 0 μ − 3 λ = 4 … ( i ) Why?
Two vectors are perpendicular exactly when their dot product is 0. So P Q → \overrightarrow{PQ} PQ 1 , − 2 , 2 1, -2, 2 1 , − 2 , 2
Multiply out each product:
( − 10 + 3 μ − λ ) ( 1 ) = − 10 + 3 μ − λ ( − 2 μ − 2 + 2 λ ) ( − 2 ) = 4 μ + 4 − 4 λ ( − 3 − 2 μ − 2 λ ) ( 2 ) = − 6 − 4 μ − 4 λ \begin{aligned} &(-10 + 3\mu - \lambda)(1) \\ &= -10 + 3\mu - \lambda \\[4pt] &(-2\mu - 2 + 2\lambda)(-2) \\ &= 4\mu + 4 - 4\lambda \\[4pt] &(-3 - 2\mu - 2\lambda)(2) \\ &= -6 - 4\mu - 4\lambda \end{aligned} ( − 10 + 3 μ − λ ) ( 1 ) = − 10 + 3 μ − λ ( − 2 μ − 2 + 2 λ ) ( − 2 ) = 4 μ + 4 − 4 λ ( − 3 − 2 μ − 2 λ ) ( 2 ) = − 6 − 4 μ − 4 λ Collect the terms:
μ : 3 + 4 − 4 = 3 λ : − 1 − 4 − 4 = − 9 number : − 10 + 4 − 6 = − 12 \begin{aligned} \mu&:\ 3 + 4 - 4 = 3 \\ \lambda&:\ -1 - 4 - 4 = -9 \\ \text{number}&:\ -10 + 4 - 6 = -12 \end{aligned} μ λ number : 3 + 4 − 4 = 3 : − 1 − 4 − 4 = − 9 : − 10 + 4 − 6 = − 12 Then divide by 3.
3 μ − 9 λ − 12 = 0 3 μ − 9 λ = 12 3 μ 3 − 9 λ 3 = 12 3 μ − 3 λ = 4 \begin{aligned} 3\mu - 9\lambda - 12 &= 0 \\ 3\mu - 9\lambda &= 12 \\ \frac{3\mu}{3} - \frac{9\lambda}{3} &= \frac{12}{3} \\ \mu - 3\lambda &= 4 \end{aligned} 3 μ − 9 λ − 12 3 μ − 9 λ 3 3 μ − 3 9 λ μ − 3 λ = 0 = 12 = 3 12 = 4 ( − 10 + 3 μ − λ ) 3 + ( − 2 μ − 2 + 2 λ ) ( − 2 ) + ( − 3 − 2 μ − 2 λ ) ( − 2 ) = 0 ⇒ 17 μ − 3 λ − 20 = 0 ⇒ 17 μ − 3 λ = 20 … ( ii ) \begin{aligned} &(-10 + 3\mu - \lambda)3 \\ &\quad + (-2\mu - 2 + 2\lambda)(-2) \\ &\quad + (-3 - 2\mu - 2\lambda)(-2) = 0 \\ \Rightarrow\ &17\mu - 3\lambda - 20 = 0 \\ \Rightarrow\ &17\mu - 3\lambda = 20 \quad\ldots(\text{ii}) \end{aligned} ⇒ ⇒ ( − 10 + 3 μ − λ ) 3 + ( − 2 μ − 2 + 2 λ ) ( − 2 ) + ( − 3 − 2 μ − 2 λ ) ( − 2 ) = 0 17 μ − 3 λ − 20 = 0 17 μ − 3 λ = 20 … ( ii ) Why?
The same with the direction 3 , − 2 , − 2 3, -2, -2 3 , − 2 , − 2
( − 10 + 3 μ − λ ) ( 3 ) = − 30 + 9 μ − 3 λ ( − 2 μ − 2 + 2 λ ) ( − 2 ) = 4 μ + 4 − 4 λ ( − 3 − 2 μ − 2 λ ) ( − 2 ) = 6 + 4 μ + 4 λ \begin{aligned} &(-10 + 3\mu - \lambda)(3) \\ &= -30 + 9\mu - 3\lambda \\[4pt] &(-2\mu - 2 + 2\lambda)(-2) \\ &= 4\mu + 4 - 4\lambda \\[4pt] &(-3 - 2\mu - 2\lambda)(-2) \\ &= 6 + 4\mu + 4\lambda \end{aligned} ( − 10 + 3 μ − λ ) ( 3 ) = − 30 + 9 μ − 3 λ ( − 2 μ − 2 + 2 λ ) ( − 2 ) = 4 μ + 4 − 4 λ ( − 3 − 2 μ − 2 λ ) ( − 2 ) = 6 + 4 μ + 4 λ Collect the terms:
μ : 9 + 4 + 4 = 17 λ : − 3 − 4 + 4 = − 3 number : − 30 + 4 + 6 = − 20 \begin{aligned} \mu&:\ 9 + 4 + 4 = 17 \\ \lambda&:\ -3 - 4 + 4 = -3 \\ \text{number}&:\ -30 + 4 + 6 = -20 \end{aligned} μ λ number : 9 + 4 + 4 = 17 : − 3 − 4 + 4 = − 3 : − 30 + 4 + 6 = − 20 ( ii ) − ( i ) : 16 μ = 16 ⇒ μ = 1 From (i): 1 − 3 λ = 4 ⇒ λ = − 1 \begin{aligned} (\text{ii}) - (\text{i}):\ &16\mu = 16 \\ \Rightarrow\ &\mu = 1 \\ \text{From (i): }&1 - 3\lambda = 4 \\ \Rightarrow\ &\lambda = -1 \end{aligned} ( ii ) − ( i ) : ⇒ From (i): ⇒ 16 μ = 16 μ = 1 1 − 3 λ = 4 λ = − 1 Solving (i) and (ii), μ = 1 \mu = 1 μ = 1 λ = − 1 \lambda = -1 λ = − 1
Why?
Both equations contain − 3 λ -3\lambda − 3 λ λ \lambda λ μ = 1 \mu = 1 μ = 1 − 3 λ = 3 -3\lambda = 3 − 3 λ = 3
( 17 μ − 3 λ ) − ( μ − 3 λ ) = 17 μ − μ − 3 λ + 3 λ = 16 μ 20 − 4 = 16 16 μ = 16 ⇒ μ = 16 16 = 1 1 − 3 λ = 4 − 3 λ = 4 − 1 = 3 λ = 3 − 3 = − 1 \begin{aligned} &(17\mu - 3\lambda) - (\mu - 3\lambda) \\ &= 17\mu - \mu - 3\lambda + 3\lambda \\ &= 16\mu \\ &20 - 4 = 16 \\[4pt] &16\mu = 16 \Rightarrow \mu = \frac{16}{16} = 1 \\[4pt] &1 - 3\lambda = 4 \\ &-3\lambda = 4 - 1 = 3 \\ &\lambda = \frac{3}{-3} = -1 \end{aligned} ( 17 μ − 3 λ ) − ( μ − 3 λ ) = 17 μ − μ − 3 λ + 3 λ = 16 μ 20 − 4 = 16 16 μ = 16 ⇒ μ = 16 16 = 1 1 − 3 λ = 4 − 3 λ = 4 − 1 = 3 λ = − 3 3 = − 1 The points at which they should be so that the distance between them is the shortest have position vectors 5 i ^ + 4 j ^ 5\hat i + 4\hat j 5 i ^ + 4 j ^ − i ^ − 2 j ^ − 3 k ^ -\hat i - 2\hat j - 3\hat k − i ^ − 2 j ^ − 3 k ^
Why?
Put λ = − 1 \lambda = -1 λ = − 1 μ = 1 \mu = 1 μ = 1
P : ( 6 + ( − 1 ) , 2 − 2 ( − 1 ) , 2 + 2 ( − 1 ) ) = ( 6 − 1 , 2 + 2 , 2 − 2 ) = ( 5 , 4 , 0 ) Q : ( − 4 + 3 ( 1 ) , − 2 ( 1 ) , − 1 − 2 ( 1 ) ) = ( − 4 + 3 , − 2 , − 1 − 2 ) = ( − 1 , − 2 , − 3 ) \begin{aligned} P&:\ \big(6 + (-1),\ 2 - 2(-1), \\ &\qquad 2 + 2(-1)\big) \\ &= (6 - 1, 2 + 2, 2 - 2) \\ &= (5, 4, 0) \\[4pt] Q&:\ \big(-4 + 3(1),\ -2(1), \\ &\qquad -1 - 2(1)\big) \\ &= (-4 + 3, -2, -1 - 2) \\ &= (-1, -2, -3) \end{aligned} P Q : ( 6 + ( − 1 ) , 2 − 2 ( − 1 ) , 2 + 2 ( − 1 ) ) = ( 6 − 1 , 2 + 2 , 2 − 2 ) = ( 5 , 4 , 0 ) : ( − 4 + 3 ( 1 ) , − 2 ( 1 ) , − 1 − 2 ( 1 ) ) = ( − 4 + 3 , − 2 , − 1 − 2 ) = ( − 1 , − 2 , − 3 ) P Q → = − 6 i ^ − 6 j ^ − 3 k ^ ∣ P Q → ∣ = 6 2 + 6 2 + 3 2 = 9 \begin{aligned} \overrightarrow{PQ} &= -6\hat i - 6\hat j - 3\hat k \\ |\overrightarrow{PQ}| &= \sqrt{6^2 + 6^2 + 3^2} = 9 \end{aligned} PQ ∣ PQ ∣ = − 6 i ^ − 6 j ^ − 3 k ^ = 6 2 + 6 2 + 3 2 = 9 Why?
P Q → \overrightarrow{PQ} PQ ( − 1 − 5 , − 2 − 4 , − 3 − 0 ) (-1 - 5, -2 - 4, -3 - 0) ( − 1 − 5 , − 2 − 4 , − 3 − 0 ) p p p q q q r r r
p 2 + q 2 + r 2 \sqrt{p^2 + q^2 + r^2} p 2 + q 2 + r 2 Here 36 + 36 + 9 = 81 36 + 36 + 9 = 81 36 + 36 + 9 = 81 81 = 9 \sqrt{81} = 9 81 = 9
( − 1 − 5 , − 2 − 4 , − 3 − 0 ) = ( − 6 , − 6 , − 3 ) ( − 6 ) 2 + ( − 6 ) 2 + ( − 3 ) 2 = 36 + 36 + 9 = 72 + 9 = 81 \begin{aligned} &(-1 - 5, -2 - 4, -3 - 0) \\ &= (-6, -6, -3) \\[4pt] &(-6)^2 + (-6)^2 + (-3)^2 \\ &= 36 + 36 + 9 \\ &= 72 + 9 = 81 \end{aligned} ( − 1 − 5 , − 2 − 4 , − 3 − 0 ) = ( − 6 , − 6 , − 3 ) ( − 6 ) 2 + ( − 6 ) 2 + ( − 3 ) 2 = 36 + 36 + 9 = 72 + 9 = 81 ∴ \therefore ∴
(b) Eliminating t t t
x 2 = y − 4 = z 4 \frac{x}{2} = \frac{y}{-4} = \frac{z}{4} 2 x = − 4 y = 4 z Why?
From x = 2 t x = 2t x = 2 t y = − 4 t y = -4t y = − 4 t z = 4 t z = 4t z = 4 t x 2 \frac{x}{2} 2 x y − 4 \frac{y}{-4} − 4 y z 4 \frac{z}{4} 4 z t t t
This is the symmetric form of a line through ( 0 , 0 , 0 ) (0, 0, 0) ( 0 , 0 , 0 )
This is the line through the origin with direction ratios 2 , − 4 , 4 2, -4, 4 2 , − 4 , 4
When t = 10 t = 10 t = 10 ( 20 , − 40 , 40 ) (20, -40, 40) ( 20 , − 40 , 40 )
20 2 + 40 2 + 40 2 = 20 × 3 = 60 km \begin{aligned} &\sqrt{20^2 + 40^2 + 40^2} \\ &= 20 \times 3 = 60 \text{ km} \end{aligned} 2 0 2 + 4 0 2 + 4 0 2 = 20 × 3 = 60 km Why?
Put t = 10 t = 10 t = 10 x = 20 x = 20 x = 20 y = − 40 y = -40 y = − 40 z = 40 z = 40 z = 40 ( x , y , z ) (x, y, z) ( x , y , z ) x 2 + y 2 + z 2 \sqrt{x^2 + y^2 + z^2} x 2 + y 2 + z 2
Take out the common factor 20:
20 2 + 40 2 + 40 2 = 20 2 + ( 2 × 20 ) 2 + ( 2 × 20 ) 2 = 20 2 ( 1 + 4 + 4 ) 20 2 ( 1 + 4 + 4 ) = 20 9 = 20 × 3 \begin{aligned} &20^2 + 40^2 + 40^2 \\ &= 20^2 + (2 \times 20)^2 + (2 \times 20)^2 \\ &= 20^2(1 + 4 + 4) \\[4pt] &\sqrt{20^2(1 + 4 + 4)} \\ &= 20\sqrt9 = 20 \times 3 \end{aligned} 2 0 2 + 4 0 2 + 4 0 2 = 2 0 2 + ( 2 × 20 ) 2 + ( 2 × 20 ) 2 = 2 0 2 ( 1 + 4 + 4 ) 2 0 2 ( 1 + 4 + 4 ) = 20 9 = 20 × 3 Check: 400 + 1600 + 1600 = 3600 400 + 1600 + 1600 = 3600 400 + 1600 + 1600 = 3600 3600 = 60 \sqrt{3600} = 60 3600 = 60
For the line, a ⃗ 1 = 20 i ^ − 10 j ^ + 40 k ^ \vec a_1 = 20\hat i - 10\hat j + 40\hat k a 1 = 20 i ^ − 10 j ^ + 40 k ^ b ⃗ = 10 i ^ − 20 j ^ + 10 k ^ \vec b = 10\hat i - 20\hat j + 10\hat k b = 10 i ^ − 20 j ^ + 10 k ^ a ⃗ 2 = 20 i ^ − 40 j ^ + 40 k ^ \vec a_2 = 20\hat i - 40\hat j + 40\hat k a 2 = 20 i ^ − 40 j ^ + 40 k ^
( a ⃗ 2 − a ⃗ 1 ) × b ⃗ = − 30 j ^ × ( 10 i ^ − 20 j ^ + 10 k ^ ) = − 300 i ^ + 300 k ^ \begin{aligned} &(\vec a_2 - \vec a_1) \times \vec b \\ &= -30\hat j \times (10\hat i - 20\hat j + 10\hat k) \\ &= -300\hat i + 300\hat k \end{aligned} ( a 2 − a 1 ) × b = − 30 j ^ × ( 10 i ^ − 20 j ^ + 10 k ^ ) = − 300 i ^ + 300 k ^ Why?
Subtract a ⃗ 1 \vec a_1 a 1 a ⃗ 2 \vec a_2 a 2
i ^ : 20 − 20 = 0 j ^ : − 40 − ( − 10 ) = − 30 k ^ : 40 − 40 = 0 \begin{aligned} \hat i&:\ 20 - 20 = 0 \\ \hat j&:\ -40 - (-10) = -30 \\ \hat k&:\ 40 - 40 = 0 \end{aligned} i ^ j ^ k ^ : 20 − 20 = 0 : − 40 − ( − 10 ) = − 30 : 40 − 40 = 0 So a ⃗ 2 − a ⃗ 1 = − 30 j ^ \vec a_2 - \vec a_1 = -30\hat j a 2 − a 1 = − 30 j ^
Cross each term using
j ^ × i ^ = − k ^ j ^ × j ^ = 0 ⃗ j ^ × k ^ = i ^ \begin{aligned} \hat j \times \hat i &= -\hat k \\ \hat j \times \hat j &= \vec 0 \\ \hat j \times \hat k &= \hat i \end{aligned} j ^ × i ^ j ^ × j ^ j ^ × k ^ = − k ^ = 0 = i ^ So − 30 j ^ × 10 i ^ = 300 k ^ -30\hat j \times 10\hat i = 300\hat k − 30 j ^ × 10 i ^ = 300 k ^ 0 ⃗ \vec 0 0 − 30 j ^ × 10 k ^ = − 300 i ^ -30\hat j \times 10\hat k = -300\hat i − 30 j ^ × 10 k ^ = − 300 i ^
− 30 j ^ × 10 i ^ = − 300 ( j ^ × i ^ ) = − 300 ( − k ^ ) = 300 k ^ − 30 j ^ × ( − 20 j ^ ) = 600 ( j ^ × j ^ ) = 600 0 ⃗ = 0 ⃗ − 30 j ^ × 10 k ^ = − 300 ( j ^ × k ^ ) = − 300 i ^ \begin{aligned} -30\hat j \times 10\hat i &= -300(\hat j \times \hat i) \\ &= -300(-\hat k) = 300\hat k \\[4pt] -30\hat j \times (-20\hat j) &= 600(\hat j \times \hat j) \\ &= 600\,\vec 0 = \vec 0 \\[4pt] -30\hat j \times 10\hat k &= -300(\hat j \times \hat k) \\ &= -300\hat i \end{aligned} − 30 j ^ × 10 i ^ − 30 j ^ × ( − 20 j ^ ) − 30 j ^ × 10 k ^ = − 300 ( j ^ × i ^ ) = − 300 ( − k ^ ) = 300 k ^ = 600 ( j ^ × j ^ ) = 600 0 = 0 = − 300 ( j ^ × k ^ ) = − 300 i ^ ∣ b ⃗ ∣ = 10 2 + 20 2 + 10 2 = 10 6 |\vec b| = \sqrt{10^2 + 20^2 + 10^2} = 10\sqrt6 ∣ b ∣ = 1 0 2 + 2 0 2 + 1 0 2 = 10 6 The distance of the point ( 20 , − 40 , 40 ) (20, -40, 40) ( 20 , − 40 , 40 )
∣ ( a ⃗ 2 − a ⃗ 1 ) × b ⃗ ∣ ∣ b ⃗ ∣ = ∣ − 300 i ^ + 300 k ^ ∣ 10 6 = 300 2 10 6 = 10 3 \begin{aligned} &\frac{|(\vec a_2 - \vec a_1) \times \vec b|}{|\vec b|} \\ &= \frac{\lvert -300\hat i + 300\hat k\rvert}{10\sqrt6} \\ &= \frac{300\sqrt2}{10\sqrt6} = 10\sqrt3 \end{aligned} ∣ b ∣ ∣ ( a 2 − a 1 ) × b ∣ = 10 6 ∣ − 300 i ^ + 300 k ^ ∣ = 10 6 300 2 = 10 3 Why?
The distance of the point with position vector a ⃗ 2 \vec a_2 a 2 r ⃗ = a ⃗ 1 + μ b ⃗ \vec r = \vec a_1 + \mu\vec b r = a 1 + μ b ( a ⃗ 2 − a ⃗ 1 ) × b ⃗ (\vec a_2 - \vec a_1) \times \vec b ( a 2 − a 1 ) × b ∣ b ⃗ ∣ |\vec b| ∣ b ∣
300 2 + 300 2 = 300 2 300 2 10 6 = 30 3 = 10 3 \begin{aligned} &\sqrt{300^2 + 300^2} = 300\sqrt2 \\ &\frac{300\sqrt2}{10\sqrt6} = \frac{30}{\sqrt3} = 10\sqrt3 \end{aligned} 30 0 2 + 30 0 2 = 300 2 10 6 300 2 = 3 30 = 10 3 In the last step, multiply top and bottom by 3 \sqrt3 3 30 3 3 = 10 3 \frac{30\sqrt3}{3} = 10\sqrt3 3 30 3 = 10 3
In full, line by line:
∣ b ⃗ ∣ 2 = 10 2 + ( − 20 ) 2 + 10 2 = 100 + 400 + 100 = 600 ∣ b ⃗ ∣ = 600 = 100 × 6 = 10 6 ( − 300 ) 2 + 0 2 + 300 2 = 2 × 300 2 = 300 2 300 2 10 6 = 30 2 2 3 = 30 3 = 30 × 3 3 × 3 = 30 3 3 = 10 3 \begin{aligned} |\vec b|^2 &= 10^2 + (-20)^2 + 10^2 \\ &= 100 + 400 + 100 = 600 \\ |\vec b| &= \sqrt{600} = \sqrt{100 \times 6} \\ &= 10\sqrt6 \\[4pt] &\sqrt{(-300)^2 + 0^2 + 300^2} \\ &= \sqrt{2 \times 300^2} = 300\sqrt2 \\[4pt] \frac{300\sqrt2}{10\sqrt6} &= \frac{30\sqrt2}{\sqrt2\,\sqrt3} \\ &= \frac{30}{\sqrt3} \\ &= \frac{30 \times \sqrt3}{\sqrt3 \times \sqrt3} \\ &= \frac{30\sqrt3}{3} = 10\sqrt3 \end{aligned} ∣ b ∣ 2 ∣ b ∣ 10 6 300 2 = 1 0 2 + ( − 20 ) 2 + 1 0 2 = 100 + 400 + 100 = 600 = 600 = 100 × 6 = 10 6 ( − 300 ) 2 + 0 2 + 30 0 2 = 2 × 30 0 2 = 300 2 = 2 3 30 2 = 3 30 = 3 × 3 30 × 3 = 3 30 3 = 10 3 ∴ \therefore ∴ 10 3 10\sqrt3 10 3