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1. Consider the point A(4,0,3)A(4,0,3) and the plane P\mathcal{P} given by \[ \mathc...
May 30, 2024
Solution by Steps
step 1
The normal vector to the plane P:2x−y+2z=5\mathcal{P}: 2x - y + 2z = 5 is given by the coefficients of xx, yy, and zz in the plane equation
step 2
Therefore, the normal vector is n=(2,−1,2)\mathbf{n} = (2, -1, 2)
# Part (b)
step 1
To find the coordinates of the point on P\mathcal{P} that is closest to A(4,0,3)A(4,0,3), we use the projection of A\mathbf{A} onto the normal vector n\mathbf{n}
step 2
The projection formula is projnA=A⋅nn⋅nn\text{proj}_{\mathbf{n}} \mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} \mathbf{n}
step 3
Calculate A⋅n=4⋅2+0⋅(−1)+3⋅2=8+6=14\mathbf{A} \cdot \mathbf{n} = 4 \cdot 2 + 0 \cdot (-1) + 3 \cdot 2 = 8 + 6 = 14
step 4
Calculate n⋅n=22+(−1)2+22=4+1+4=9\mathbf{n} \cdot \mathbf{n} = 2^2 + (-1)^2 + 2^2 = 4 + 1 + 4 = 9
step 5
Therefore, projnA=149(2,−1,2)=(289,−149,289)\text{proj}_{\mathbf{n}} \mathbf{A} = \frac{14}{9} (2, -1, 2) = \left(\frac{28}{9}, -\frac{14}{9}, \frac{28}{9}\right)
step 6
The coordinates of the point on P\mathcal{P} closest to AA are (289,−149,289)\left(\frac{28}{9}, -\frac{14}{9}, \frac{28}{9}\right)
Answer
The normal vector to the plane is n=(2,−1,2)\mathbf{n} = (2, -1, 2). The coordinates of the point on P\mathcal{P} closest to AA are (289,−149,289)\left(\frac{28}{9}, -\frac{14}{9}, \frac{28}{9}\right).
Key Concept
Normal vector and projection
Explanation
The normal vector is derived from the coefficients of the plane equation. The closest point on the plane to a given point is found using the projection of the point onto the normal vector.
Question 2 # Part (a)
step 1
To find u×v\mathbf{u} \times \mathbf{v}, we use the cross product formula
step 2
u=(4,1,−1)\mathbf{u} = (4, 1, -1) and v=(2,2,1)\mathbf{v} = (2, 2, 1)
step 3
u×v=∣iamp;jamp;k4amp;1amp;−12amp;2amp;1∣=(1⋅1−(−1)⋅2)i−(4⋅1−(−1)⋅2)j+(4⋅2−1⋅2)k=(3,−6,6)\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 1 & -1 \\ 2 & 2 & 1 \end{vmatrix} = (1 \cdot 1 - (-1) \cdot 2) \mathbf{i} - (4 \cdot 1 - (-1) \cdot 2) \mathbf{j} + (4 \cdot 2 - 1 \cdot 2) \mathbf{k} = (3, -6, 6)
# Part (b)
step 1
To find the point-normal equation of the plane passing through D(1,0,0)D(1,0,0) and parallel to u\mathbf{u} and v\mathbf{v}, we use the normal vector u×\mathbf{u} \times \mathbf{v} = (3, -6, 6)$
step 2
The point-normal form of the plane equation is n⋅(r−r0)=0\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0, where r0=(1,0,0)\mathbf{r_0} = (1,0,0)
step 3
Substituting n=(3,−6,6)\mathbf{n} = (3, -6, 6) and r0=(1,0,0)\mathbf{r_0} = (1,0,0), we get 3(x−1)−6(y−0)+6(z−0)=03(x - 1) - 6(y - 0) + 6(z - 0) = 0
step 4
Simplifying, we get the Cartesian equation 3x−6y+6z=33x - 6y + 6z = 3
Answer
u×v=(3,−6,6)\mathbf{u} \times \mathbf{v} = (3, -6, 6). The Cartesian equation of the plane is 3x−6y+6z=33x - 6y + 6z = 3.
Key Concept
Cross product and plane equation
Explanation
The cross product of two vectors gives a vector orthogonal to both. The point-normal form of a plane equation uses a point on the plane and a normal vector.
Question 3
step 1
To find a vector r\mathbf{r} orthogonal to both p=(2,−1,2)\mathbf{p} = (2, -1, 2) and q=(1,0,−2)\mathbf{q} = (1, 0, -2), we use the cross product p×q\mathbf{p} \times \mathbf{q}
step 2
p×q=∣iamp;jamp;k2amp;−1amp;21amp;0amp;−2∣=(−1⋅−2−2⋅0)i−(2⋅−2−2⋅1)j+(2⋅0−(−1)⋅1)k=(2,−2,1)\mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix} = (-1 \cdot -2 - 2 \cdot 0) \mathbf{i} - (2 \cdot -2 - 2 \cdot 1) \mathbf{j} + (2 \cdot 0 - (-1) \cdot 1) \mathbf{k} = (2, -2, 1)
step 3
The volume of the parallelepiped spanned by p\mathbf{p}, q\mathbf{q}, and r\mathbf{r} is given by the scalar triple product ∣p⋅(q×r)∣|\mathbf{p} \cdot (\mathbf{q} \times \mathbf{r})|
step 4
Calculate q×r=∣iamp;jamp;k1amp;0amp;−22amp;−2amp;1∣=(0⋅1−(−2)⋅−2)i−(1⋅1−(−2)⋅2)j+(1⋅−2−0⋅2)k=(−4,5,−2)\mathbf{q} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -2 \\ 2 & -2 & 1 \end{vmatrix} = (0 \cdot 1 - (-2) \cdot -2) \mathbf{i} - (1 \cdot 1 - (-2) \cdot 2) \mathbf{j} + (1 \cdot -2 - 0 \cdot 2) \mathbf{k} = (-4, 5, -2)
step 5
Calculate p⋅(−4,5,−2)=2⋅−4+(−1)⋅5+2⋅−2=−8−5−4=−17\mathbf{p} \cdot (-4, 5, -2) = 2 \cdot -4 + (-1) \cdot 5 + 2 \cdot -2 = -8 - 5 - 4 = -17
step 6
The volume is ∣−17∣=17| -17 | = 17
Answer
r=(2,−2,1)\mathbf{r} = (2, -2, 1). The volume of the parallelepiped is 17 cubic units.
Key Concept
Cross product and scalar triple product
Explanation
The cross product of two vectors gives a vector orthogonal to both. The scalar triple product gives the volume of the parallelepiped formed by three vectors.
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