Matrices & Vectors - Multivariable Calculus

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Question

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

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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

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

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Question

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

Tap to reveal answer

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

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

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Question

Find the equation of the tangent plane to at .

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Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(0,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(0,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

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Question

Find the equation of the tangent plane to at .

Tap to reveal answer

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(1,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(1,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

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Question

Find the equation of the tangent plane to at .

Tap to reveal answer

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(0,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(0,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

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Question

Find the equation of the tangent plane to at .

Tap to reveal answer

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(1,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(1,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

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Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

← Didn't Know|Knew It →

Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

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

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Question

Find the equation of the tangent plane to at .

Tap to reveal answer

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(0,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(0,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

← Didn't Know|Knew It →

Question

Find the equation of the tangent plane to at .

Tap to reveal answer

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

,

$f_$x(x,y)=\frac{12x^2$$}{4x^3$$+10y^2$}$$ , $f_$x(1,5)=\frac{12(1)^2$$}{4(1)^3$$+10(5)^2$}$=\frac{12}{254}$=\frac{6}{127}$$

$f_$y(x,y)=\frac{20y}{4x^3$$+10y^2$}$$ , $f_$y(1,5)=\frac{20(5)}{4(1)^3$$+10(5)^2$}$=\frac{100}{254}$=\frac{50}{127}$$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)$

$z=\frac{6}{127}$(x-1)+\frac{50}{127}$(y-5)+\ln(254)$

$z=\frac{6}{127}$x-$\frac{6}{127}$+\frac{50}{127}$y-$\frac{250}{127}$+\ln(254)$

$z=\frac{6}{127}$x+\frac{50}{127}$y-$\frac{256}{127}$+\ln(254)$

← Didn't Know|Knew It →

Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

← Didn't Know|Knew It →

Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

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

← Didn't Know|Knew It →

Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

← Didn't Know|Knew It →

Question

Let , and .

Find $u\times v$ .

Tap to reveal answer

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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