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To derive a numerical model for fluid behaviour in pipes, vectors are used. It...
May 16, 2024
To derive a numerical model for fluid behaviour in pipes, vectors are used. It is given that points A, B and C have coordinates (2, โ€“1, โ€“5), (0, 5, โ€“ 9) and (9, 2, 3) respectively. The line l has equation r= (2i -j -5k)+๐›ฝ(๐‘– โˆ’ 3๐‘— + 2๐‘˜) (a) Verify that the point B lies on the line l. (b) Find the vector equation of BC. (c) The point D is such that vector AD= 2BC. (i) Show that D has coordinates (20, โ€“7, 19). (ii) The point P lies on l where ฮป = p. The line PD is perpendicular to l. Find the value of p. (d) Find the cartesian equation of: i. Line l. ii. Line AC.
Solution by Steps
step 1
The line l l has the equation r=(2iโˆ’jโˆ’5k)+ฮฒ(iโˆ’3j+2k) \mathbf{r} = (2\mathbf{i} - \mathbf{j} - 5\mathbf{k}) + \beta(\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})
step 2
Substitute the coordinates of point B, (0,5,โˆ’9) (0, 5, -9) , into the line equation to find ฮฒ \beta
step 3
Set up the equations: {2+ฮฒ=0โˆ’1โˆ’3ฮฒ=5โˆ’5+2ฮฒ=โˆ’9 \begin{cases} 2 + \beta = 0 \\ -1 - 3\beta = 5 \\ -5 + 2\beta = -9 \end{cases}
step 4
Solve for ฮฒ \beta : ฮฒ=โˆ’2 \beta = -2
step 5
Verify the solution by substituting ฮฒ=โˆ’2 \beta = -2 back into the line equation: r=(2iโˆ’jโˆ’5k)+(โˆ’2)(iโˆ’3j+2k)=0i+5jโˆ’9k \mathbf{r} = (2\mathbf{i} - \mathbf{j} - 5\mathbf{k}) + (-2)(\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) = 0\mathbf{i} + 5\mathbf{j} - 9\mathbf{k}
step 6
Since the coordinates match, point B lies on the line l l
(b) Find the vector equation of BC.
step 1
The coordinates of points B and C are (0,5,โˆ’9) (0, 5, -9) and (9,2,3) (9, 2, 3) respectively
step 2
The vector BC \mathbf{BC} is given by: BC=Cโˆ’B=(9โˆ’0)i+(2โˆ’5)j+(3+9)k=9iโˆ’3j+12k \mathbf{BC} = \mathbf{C} - \mathbf{B} = (9 - 0)\mathbf{i} + (2 - 5)\mathbf{j} + (3 + 9)\mathbf{k} = 9\mathbf{i} - 3\mathbf{j} + 12\mathbf{k}
(c) The point D is such that vector AD = 2BC. # (i) Show that D has coordinates (20, โ€“7, 19).
step 1
The coordinates of point A are (2,โˆ’1,โˆ’5) (2, -1, -5)
step 2
The vector AD=2BC \mathbf{AD} = 2\mathbf{BC} : AD=2(9iโˆ’3j+12k)=18iโˆ’6j+24k \mathbf{AD} = 2(9\mathbf{i} - 3\mathbf{j} + 12\mathbf{k}) = 18\mathbf{i} - 6\mathbf{j} + 24\mathbf{k}
step 3
The coordinates of point D are: D=A+AD=(2,โˆ’1,โˆ’5)+(18,โˆ’6,24)=(20,โˆ’7,19) \mathbf{D} = \mathbf{A} + \mathbf{AD} = (2, -1, -5) + (18, -6, 24) = (20, -7, 19)
# (ii) The point P lies on l where ฮป=p \lambda = p . The line PD is perpendicular to l. Find the value of p.
step 1
The direction vector of line l l is d=iโˆ’3j+2k \mathbf{d} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k}
step 2
The coordinates of point P are: P=(2+p,โˆ’1โˆ’3p,โˆ’5+2p) \mathbf{P} = (2 + p, -1 - 3p, -5 + 2p)
step 3
The vector PD \mathbf{PD} is: PD=(20โˆ’(2+p))i+(โˆ’7โˆ’(โˆ’1โˆ’3p))j+(19โˆ’(โˆ’5+2p))k=(18โˆ’p)i+(โˆ’6+3p)j+(24โˆ’2p)k \mathbf{PD} = (20 - (2 + p))\mathbf{i} + (-7 - (-1 - 3p))\mathbf{j} + (19 - (-5 + 2p))\mathbf{k} = (18 - p)\mathbf{i} + (-6 + 3p)\mathbf{j} + (24 - 2p)\mathbf{k}
step 4
Since PD \mathbf{PD} is perpendicular to d \mathbf{d} , their dot product is zero: (18โˆ’p)โ‹…1+(โˆ’6+3p)โ‹…(โˆ’3)+(24โˆ’2p)โ‹…2=0 (18 - p) \cdot 1 + (-6 + 3p) \cdot (-3) + (24 - 2p) \cdot 2 = 0
step 5
Simplify and solve for p p : 18โˆ’p+18โˆ’9p+48โˆ’4p=0โ€…โ€ŠโŸนโ€…โ€Š84โˆ’14p=0โ€…โ€ŠโŸนโ€…โ€Šp=6 18 - p + 18 - 9p + 48 - 4p = 0 \implies 84 - 14p = 0 \implies p = 6
(d) Find the cartesian equation of: # i. Line l.
step 1
The parametric equations of line l l are: x=2+ฮฒ,y=โˆ’1โˆ’3ฮฒ,z=โˆ’5+2ฮฒ x = 2 + \beta, \quad y = -1 - 3\beta, \quad z = -5 + 2\beta
step 2
Eliminate ฮฒ \beta to find the cartesian equation: ฮฒ=xโˆ’2,ฮฒ=y+1โˆ’3,ฮฒ=z+52 \beta = x - 2, \quad \beta = \frac{y + 1}{-3}, \quad \beta = \frac{z + 5}{2}
step 3
Equate the expressions for ฮฒ \beta : xโˆ’2=y+1โˆ’3=z+52 x - 2 = \frac{y + 1}{-3} = \frac{z + 5}{2}
# ii. Line AC.
step 1
The coordinates of points A and C are (2,โˆ’1,โˆ’5) (2, -1, -5) and (9,2,3) (9, 2, 3) respectively
step 2
The direction vector AC \mathbf{AC} is: AC=(9โˆ’2)i+(2+1)j+(3+5)k=7i+3j+8k \mathbf{AC} = (9 - 2)\mathbf{i} + (2 + 1)\mathbf{j} + (3 + 5)\mathbf{k} = 7\mathbf{i} + 3\mathbf{j} + 8\mathbf{k}
step 3
The parametric equations of line AC AC are: x=2+7t,y=โˆ’1+3t,z=โˆ’5+8t x = 2 + 7t, \quad y = -1 + 3t, \quad z = -5 + 8t
step 4
Eliminate t t to find the cartesian equation: t=xโˆ’27,t=y+13,t=z+58 t = \frac{x - 2}{7}, \quad t = \frac{y + 1}{3}, \quad t = \frac{z + 5}{8}
step 5
Equate the expressions for t t : xโˆ’27=y+13=z+58 \frac{x - 2}{7} = \frac{y + 1}{3} = \frac{z + 5}{8}
Answer
(a) Point B lies on the line l l .
(b) The vector equation of BC \mathbf{BC} is 9iโˆ’3j+12k 9\mathbf{i} - 3\mathbf{j} + 12\mathbf{k} .
(c)(i) The coordinates of point D are (20,โˆ’7,19) (20, -7, 19) .
(c)(ii) The value of p p is 6 6 .
(d)(i) The cartesian equation of line l l is xโˆ’2=y+1โˆ’3=z+52 x - 2 = \frac{y + 1}{-3} = \frac{z + 5}{2} .
(d)(ii) The cartesian equation of line AC AC is xโˆ’27=y+13=z+58 \frac{x - 2}{7} = \frac{y + 1}{3} = \frac{z + 5}{8} .
Key Concept
Vector equations and parametric forms of lines in 3D space.
Explanation
The problem involves verifying points on a line, finding vector equations, and converting parametric equations to cartesian form.
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