AP Calculus BC - Equations of Lines and Planes

Write down the equation of the line in vector form that passes through the points , and .

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10-6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2-5t, 10+6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10+6t, -6-2t\right \rangle$

$\vec{r}=\left \langle 2+5t, 1+t, -6-20t\right \rangle$

Explanation

Remember the general equation of a line in vector form:

$\vec{r}=r_0+t\vec{v}=\left \langle x_0,y_0,z_0\right \rangle+t\left \langle a,b,c\right \rangle$ , where $\left \langle x_0,y_0,z_0\right \rangle$ is the starting point, and $\left \langle a,b,c\right \rangle$ is the difference between the start and ending points.

Lets apply this to our problem.

$\vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle$

Distribute the

$\vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle$

Now we simply do vector addition to get

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

Determine the equation of the plane given by the following two vectors and the point :

$\left \langle 3, 1, 5\right \rangle, \left \langle 2, 1, 1\right \rangle$

Explanation

The equation of a plane is given by

where the normal vector is given by $\vec{n}=\left \langle a,b,c\right \rangle$ and a point on the plane denoted

To find the normal vector to the plane, we must take the cross product of the two vectors.

We must write the determinant in order to take the cross product:

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \ 3&1 &5 \ 2&1 &1 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

$\hat{i}+3\hat{k}+10\hat{j}-(2\hat{k}+5\hat{i}+3\hat{j})= -4\hat{i}+7\hat{j}+\hat{k}=\left \langle -4, 7, 1\right \rangle$

Now that we have the normal vector and a point on the plane, we plug everything into the equation:

which simplifies to

Find the approximate angle between the planes , and .

None of the other answers.

Explanation

Finding the angle between two planes requires us to find the angle between their normal vectors.

To obtain normal vectors, we simply take the coefficients in front of .

$\mathbf{n_1} = <4,-3,-2>$

$\mathbf{n_2} = <12,2,-7>$

The (acute) angle between any two vector is

$\theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|})$ ,

Substituting, we have

$\theta = \cos^{-1}(\frac{<4,-3,-2> \cdot <12,2,-7>}{\sqrt{16+9+4}\sqrt{144+4+49}})$

$\theta \approx \cos^{-1}(\frac{56}{75.584})$

$\theta \approx 42.192$ .

Determine the equation of the plane containing the three points

Explanation

The equation of a plane is defined as

$n\bullet <x-x_0,y-y_0,z-z_0>=0$

where

is the normal vector of the plane.

To find the normal vector, we first get two vectors on the plane

and

and find their cross product.

The cross product is defined as the determinant of the matrix

$\begin{vmatrix} i& j& k\ 5& 4& 7\ -2&5 &6 \end{vmatrix}$

Which is

$=det(\begin{vmatrix} 4& 7\5 &6 \end{vmatrix})i-det(\begin{vmatrix} 5& 7\-2 &6 \end{vmatrix})j+det(\begin{vmatrix} 5& 4\ -2&5 \end{vmatrix})k$

Which tells us the normal vector is

Using the point

and the normal vector to find the equation of the plane yields

$<-11,-44,33>\bullet <x-0,y-0,z-0> = 0$

$<-11,-44,33>\bullet <x,y,z> = 0$

Simplified gives the equation of the plane

Find the unit tangent vector given by the curve

$\vec{r}(t)=(te^t)\hat{i}+t^2\hat{j}+t^3\hat{k}$

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(2t+t^2)+4t^2+9t^4}}$

$\frac{(2e^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

$(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}$

Explanation

To find the unit tangent vector for a given curve, we must find the derivative of each of the components of the curve, and then divide this vector - the tangent vector - by the magnitude of the tangent vector itself.

To start, let's find the derivatives:

$\vec{r}'(t)=(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}$

We used the following rules to find them:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$

Next, we must find the magnitude of the tangent vector, by taking the square root of the sum of the squares the x, y, and z components. Note that the vector we were given was in unit vector notation, and i, j, and k are each equal to one unit, so they don't impact the magnitude of the vector (they simply indicate direction).

$\left | \vec{r}'(t)\right |=\sqrt{(e^t+te^t)^2+(2t)^2+(3t^2)^2}=\sqrt{e^{2t}+2te^{2t}+t^2e^{2t}+4t^2+9t^4}$

Finally, divide the tangent vector by its magnitude:

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

Find the unit tangent vector to the curve given by

$\vec{r}(t)=3t\hat{i}+e^t\hat{j}+t^3\hat{k}$

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

$\frac{3\hat{i}+e^t\hat{j}+9t^4\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{3+e^{t}+3t^2}}$

${3\hat{i}+e^t\hat{j}+3t^2\hat{k}}$

Explanation

To start, let's find the derivatives:

$\vec{r}(t)=3\hat{i}+e^t\hat{j}+3t^2\hat{k}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$

$\left |\vec{r}(t) \right | = \sqrt{3^2+(e^t)^2+(3t^2)^2}=\sqrt{9+e^{2t}+9t^4}$

Finally, divide the vector by its magnitude:

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

Find the angle (in degrees) between the planes ,

Explanation

A quick way to notice the answer is is to notice the planes are parallel (They only differ by the constant on the right side).

Typically though, to find the angle between two planes, we find the angle between their normal vectors.

A vector normal to the first plane is $\mathbf{n}_1 = <1,1,1>$

A vector normal to the second plane is $\mathbf{n}_2 = <1,1,1>$

Then using the formula for the angle between vectors, $\theta = \cos^{-1} (\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|})$ , we have

$\theta = \cos^{-1} (\frac{<1,1,1> \cdot <1,1,1>}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+1^2+1^2}}) = \cos^{-1}(1) = 0$ .

Find the equation of the plane that contains the point , and has a normal vector $n=\left \langle 3,2,5 \right \rangle$ .

Explanation

Using the formula for a plane , where the point given is and the normal vector is $\left \langle A,B,C\right \rangle$ . Plugging in the known values, you get . Manipulating this equation through algebra gives you the answer