Equations of Lines and Planes - AP Calculus BC

Card 0 of 340

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

Find the unit tangent vector given by the curve

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

Answer

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:

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

Compare your answer with the correct one above

Question

Find the approximate angle between the planes , and .

Answer

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

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

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

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

$\theta \approx 42.192$ .

Compare your answer with the correct one above

Question

Find the point of intersection of the plane and the line described by

Answer

Substituting the components of the line into those of the plane, we have

$t = \frac{1}{2}$

Substituting this value of back into the components of the line gives us

$(5,-\frac{1}{2},-\frac{1}{2})$ .

Compare your answer with the correct one above

Question

Find the angle (in degrees) between the planes ,

Answer

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

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

Compare your answer with the correct one above

Question

Determine the equation of the plane that contains the following points.

Answer

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

$\begin{vmatrix} i&j &k \ 5&4 &0 \ -3&-7 &0 \end{vmatrix}$

Which is

$=det(\begin{vmatrix} 4&0 \ -7&0 \end{vmatrix})i-det(\begin{vmatrix} 5&0 \ -3&0 \end{vmatrix})j+det(\begin{vmatrix} 5&4 \ -3&-7 \end{vmatrix})k$

Which tells us the normal vector is

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

$<0,0,-23>\bullet <x-0,y-0,z-0>=0$

$<0,0,-23>\bullet <x,y,z>=0$

Simplified gives the equation of the plane

Compare your answer with the correct one above

Question

Find the equation of the plane containing the points

Answer

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

$\begin{vmatrix} i&j &k \ 0&1 &5 \ 0&-2 &-4 \end{vmatrix}$

Which is

$=det(\begin{vmatrix}1&5 \ -2&-4 \end{vmatrix})i-det(\begin{vmatrix} 0&5 \ 0&-4 \end{vmatrix})j+det(\begin{vmatrix} 0&1 \ 0&-2 \end{vmatrix})k$

Which tells us the normal vector is

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

$<6,0,0>\bullet <x-0,y-0,z-0>=0$

$<6,0,0>\bullet <x,y,z>=0$

Simplified gives the equation of the plane

Compare your answer with the correct one above

Question

Which of the following is an equation of a plane parallel to the plane ?

Answer

Planes that are parallel to each other only differ (if at all) by the constant on the right-hand side (when both sides are simplified). Since has the same coefficients as the given plane, they are parallel to each other.

Compare your answer with the correct one above

Question

Find a parametric representation of the curve of intersection of the cylinder and the plane .

Answer

We can begin by rewriting the expression for the cylinder as follows

$9x^2 + y^2 = 9 \Rightarrow x^2 + \left( \frac{y}{3} \right) ^2 = 1$ .

This tells us that $x = \cos t, y = 3 \sin t$ . Plugging this back into the equation for the plane to find $z = 7 - x - y = 7 - \cos t - 3 \sin t$ .

This gives us the representation of the curve of intersection as

$\vec{r}(t)= \left< \cos t, 3 \sin t, 7-\cos t - 3 \sin t\right>$ .

Compare your answer with the correct one above

Question

Find the equation of the plane containing the following points

Answer

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

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

Which is

$=det(\begin{vmatrix} -2&-1 \ -1&-3 \end{vmatrix})i-det(\begin{vmatrix} -1&-1 \ 3&-3 \end{vmatrix})j+det(\begin{vmatrix} -1&-2 \ 3&-1 \end{vmatrix})k$

Which tells us the normal vector is

Using the point

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

$<5,-6,7>\bullet<x-2,y-3,z-4>=0$

Simplified gives the equation of the plane

Compare your answer with the correct one above

Question

Find the angle in degrees between the planes and

Answer

To find the angle between the planes, we find the angle between their normal vectors.

We have

$\mathbf{n}_1 = <1,5,-1>$ , for the first plane, and

$\mathbf{n}_2 = <1,0,-1>$ , for the second plane.

The angle between these two vectors is

$\theta = \cos^{-1}(\frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|})$

$= \cos^{-1}(\frac{<1,5,-1> \cdot <1,0,-1>}{\sqrt{1^2+(-5)^2+(-1)^2}\sqrt{1^2+0^2+(-1)^2}})$

$\approx \cos^{-1}(\frac{2}{7.34847}) \approx 74.207$

Compare your answer with the correct one above

Question

Find the equation of a plane that passes through and .

Answer

Take the cross product of the given points. This will result in an orthogonal vector to both and , whose components will be the corresponding direction ratios.

$(1,2,3)\times(3,4,5)=(-2,4,-2)$

Equation of the plane is

Now plug in or , you will have the answer

$-2(x-1)+4(y-2)-2(z-3)=0\-2x+4y-2z+2-8+6=0\-2x+4y-2z=0\-x+2y-z=0$

Compare your answer with the correct one above

Question

Find the unit tangent vector to the curve given by

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

Answer

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:

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

The derivatives were found using the following rules:

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

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).

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

Finally, divide the vector by its magnitude:

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

Compare your answer with the correct one above

Question

Find the unit tangent vector to the given curve:

$\vec{r}=19t\hat{i}+\cos(t)\hat{j}+\sin(t)\hat{k}$

Answer

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:

$\vec{r}'(t)=19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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).

$\left | \vec{r}'(t)\right |=\sqrt{(19)^2+(-\sin(t))^2+(\cos(t))^2}=\sqrt{19^2+1}=\sqrt{382}$

Finally, divide the tangent vector by its magnitude:

$\frac{19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

Compare your answer with the correct one above

Question

Find the equation of the plane containing the points

Answer

To find the equation for the plane, we must first find two vectors contained within the plane:

$\vec{AB}=\left \langle 0, 5, 0\right \rangle, \vec{BC}=\left \langle 3, 2, 0\right \rangle$

We found these by subtracting the terminal point from the initial point for each vector (the terminal points being B and C, and the initial points A and B, respectively).

Now, we can write the determinant in order to take the cross product of the two vectors:

$\vec{AB}X\vec{BC}=\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \ 0&5 &0 \ 3& 2& 0 \end{vmatrix}$

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

Next, we must find the cross product of the two vectors, in order to determine the normal vector to the plane. 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:

$0+0+0-(15\hat{k}+0+0)=-15\hat{k}$

Now, we write out the equation of a plane, given by

, where a, b, and c are the elements of the normal vector $\left \langle a, b, c\right \rangle$ , and is any point on the plane.

Our normal vector is equal to $\left \langle 0, 0, 15\right \rangle$ , and we can pick point A, so the equation of the plane is

which becomes

This answer should make sense because none of the points have a z component, so they are all on the x-y plane, thus the only plane containing all three is the one where z has no value.

Compare your answer with the correct one above

Question

Find the equation of the plane containing the point and the normal vector given by $5\hat{j}+10\hat{k}$

Answer

The equation of a plane is given by:

, where a, b, and c are the elements of the normal vector $\left \langle a, b, c\right \rangle$ , and is any point on the plane.

Using the above formula, we get

which becomes

Compare your answer with the correct one above

Question

Find the unit tangent vector to the curve:

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

Answer

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:

$\vec{r}'(t)=\sec^2(t)\hat{i}+0\hat{j}+2e^{2t}\hat{k}$

We used the following rules to find the derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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).

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

Finally, divide the tangent vector by its magnitude:

$\frac{\sec^2(t)\hat{i}+2e^{2t}\hat{k}}{\sqrt{\sec^4(t)+4e^{4t}}}$

Compare your answer with the correct one above

Question

Find the equation of the plane containing the points

Answer

To find the equation of a plane, we need a point on the plane and the normal vector to the plane, given by the cross product of two vectors on the plane.

To start, let's find these two vectors:

$\vec{AB}=\left \langle 0, 6, 0\right \rangle, \vec{BC}=\left \langle 0, -3, 3\right \rangle$

We found these by subtracting terminal points (B, C) and initial points (A, B, respectively).

Now, we can write the determinant in order to take the cross product of the two vectors:

$\vec{AB}X\vec{BC}=\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \ 0&6 &0 \ 0& -3& 3 \end{vmatrix}$

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:

$18\hat{i}+0+0-(0+0+0)=18\hat{i}$

The equation of a plane is given by:

, where a, b, and c are the elements of the normal vector $\left \langle a, b, c\right \rangle$ , and is any point on the plane.

Our normal vector is $\left \langle 18, 0, 0\right \rangle$ , and the point we can choose is . Using the above formula, we get

Compare your answer with the correct one above

Question

Write the equation of the line - in vector form - passing through the two points

Answer

To find the equation of the line, we first must find the vector passing between the two points:

$\vec{v}=(1, -3, -5)$

This was found by .

Next, we pick one point, and using the formula

$\vec{r}=\left \langle x_0, y_0, z_0\right \rangle+t\cdot \vec{v}$

we get our line:

$\vec{r}=\left \langle 0, 3, 5\right \rangle+t\left \langle 1, -3, -5\right \rangle=\left \langle t, 3-3t, 5-5t\right \rangle$

Compare your answer with the correct one above

Question

Write the equation for the plane containing the points

Answer

To find the equation of the plane, we need a point on the plane, and the normal vector to the plane, which is given by the cross product of two vectors in the plane.

To start, let's find these vectors:

$\vec{AB}=\left \langle -2, -3, -4\right \rangle, \vec{BC}=\left \langle 3, -1, -4\right \rangle$

We found them by subtracting the terminal point (B, C) and the initial point (A, B, respectively).

Now, we can write the determinant in order to take the cross product of the two vectors:

$\vec{AB}X\vec{BC}=\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \ -2& -3&-4 \ 3& -1&-4 \end{vmatrix}$

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:

$12\hat{i}+2\hat{k}-12\hat{j}-(-9\hat{k}+4\hat{i}+8\hat{j})=8\hat{i}-20\hat{j}+11\hat{k}$

The equation of a plane is given by

, where $\left \langle a, b, c\right \rangle$ is the normal vector and is any point on the plane.

Using our normal vector $\left \langle 8, -20, 11\right \rangle$ and the point , we get the equation

Compare your answer with the correct one above

Tap the card to reveal the answer