A plane through the origin is perpendicular to the plane 2x−y−z=5 and parallel to the line joining the points (1,2,3) and (4,−1,2). Find the equation of the plane.

st3he1d0t
2022-10-05
Answered

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vakleraarrc

Answered 2022-10-06
Author has **6** answers

We know that the required plane's normal is perpendicular to both the given plane's normal (2,−1,−1) and the line's direction vector (3,−3,−1). We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:

$(2,-1,-1)\times (3,-3,1)=\dots $

$(2,-1,-1)\times (3,-3,1)=\dots $

Riya Andrews

Answered 2022-10-07
Author has **4** answers

You could as well find the parametric form of the plane:

$x=\lambda \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)+\mu \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right),\lambda ,\mu \in \mathbb{R}$

where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by

$\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\times \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)=\left(\begin{array}{c}-4\\ -5\\ -3\end{array}\right)$

from what You get the coordinate equation of the plane:

$4x+5y+3z=0.$

$x=\lambda \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)+\mu \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right),\lambda ,\mu \in \mathbb{R}$

where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by

$\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\times \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)=\left(\begin{array}{c}-4\\ -5\\ -3\end{array}\right)$

from what You get the coordinate equation of the plane:

$4x+5y+3z=0.$

asked 2021-05-29

Find the vector and parametric equations for the line segment connecting P to Q.

P(0, - 1, 1), Q(1/2, 1/3, 1/4)

asked 2021-05-17

Find the scalar and vector projections of b onto a.

$a=(4,7,-4),b=(3,-1,1)$

asked 2021-02-11

Let F be a fixed 3x2 matrix, and let H be the set of all matrices A in $M}_{2\times 4$ with the property that FA = 0 (the zero matrix in ${M}_{3\times 4})$ . Determine if H is a subspace of $M}_{2\times 4$

asked 2021-05-29

Which of the following expressions are meaningful? Which are meaningless? Explain.

a)$(a\cdot b)\cdot c$

$(a\cdot b)\cdot c$ has ? because it is the dot product of ?.

b)$(a\cdot b)c$

$(a\cdot b)c$ has ? because it is a scalar multiple of ?.

c)$|a|(b\cdot c)$

$|a|(b\cdot c)$ has ? because it is the product of ?.

d)$a\cdot (b+c)$

$a\cdot (b+c)$ has ? because it is the dot product of ?.

e)$a\cdot b+c$

$a\cdot b+c$ has ? because it is the sum of ?.

f)$|a|\cdot (b+c)$

$|a|\cdot (b+c)$ has ? because it is the dot product of ?.

a)

b)

c)

d)

e)

f)

asked 2022-07-17

I'm trying to find the magnitude of a vector $\dot{\overrightarrow{r}}(t)$ by two different methods, but get different results.

Let $\overrightarrow{r}(t)=x(t)\hat{x}+y(t)\hat{y}$. The magnitude of this vector is:

$||\overrightarrow{r}(t)||=\sqrt{x(t{)}^{2}+y(t{)}^{2}}$

The time derivative of $\overrightarrow{r}(t)$ must then be

$\dot{\overrightarrow{r}}(t)=\dot{x}(t)\hat{\dot{x}}+\dot{y}(t)\hat{\dot{y}}$

Now, using the same formula for the magnitude as before yields:

$||\dot{\overrightarrow{r}}(t)||=\sqrt{\dot{x}(t{)}^{2}+\dot{y}(t{)}^{2}}$

But when I try to find the magnitude of $\dot{\overrightarrow{r}}(t)$ by differentiating $||\overrightarrow{r}(t)||$ with respect to time, the chain rule gives me:

$\begin{array}{rl}\frac{d}{dt}||\overrightarrow{r}||& =\frac{\mathrm{\partial}||\overrightarrow{r}||}{\mathrm{\partial}x}\frac{dx}{dt}+\frac{\mathrm{\partial}||\overrightarrow{r}||}{\mathrm{\partial}y}\frac{dy}{dt}\\ & =\frac{2x}{2\sqrt{{x}^{2}+{y}^{2}}}\dot{x}+\frac{2y}{2\sqrt{{x}^{2}+{y}^{2}}}\dot{y}\\ & =\dot{x}\frac{x}{||r||}+\dot{y}\frac{y}{||r||}\ne \sqrt{{\dot{x}}^{2}+{\dot{y}}^{2}}\end{array}$

Where it is implicit that all the functions are evaluated in t. Have I made an algebraic mistake, or is it not possible to find $||\dot{\overrightarrow{r}}(t)||$ by differentiation? - and if so, why?

Let $\overrightarrow{r}(t)=x(t)\hat{x}+y(t)\hat{y}$. The magnitude of this vector is:

$||\overrightarrow{r}(t)||=\sqrt{x(t{)}^{2}+y(t{)}^{2}}$

The time derivative of $\overrightarrow{r}(t)$ must then be

$\dot{\overrightarrow{r}}(t)=\dot{x}(t)\hat{\dot{x}}+\dot{y}(t)\hat{\dot{y}}$

Now, using the same formula for the magnitude as before yields:

$||\dot{\overrightarrow{r}}(t)||=\sqrt{\dot{x}(t{)}^{2}+\dot{y}(t{)}^{2}}$

But when I try to find the magnitude of $\dot{\overrightarrow{r}}(t)$ by differentiating $||\overrightarrow{r}(t)||$ with respect to time, the chain rule gives me:

$\begin{array}{rl}\frac{d}{dt}||\overrightarrow{r}||& =\frac{\mathrm{\partial}||\overrightarrow{r}||}{\mathrm{\partial}x}\frac{dx}{dt}+\frac{\mathrm{\partial}||\overrightarrow{r}||}{\mathrm{\partial}y}\frac{dy}{dt}\\ & =\frac{2x}{2\sqrt{{x}^{2}+{y}^{2}}}\dot{x}+\frac{2y}{2\sqrt{{x}^{2}+{y}^{2}}}\dot{y}\\ & =\dot{x}\frac{x}{||r||}+\dot{y}\frac{y}{||r||}\ne \sqrt{{\dot{x}}^{2}+{\dot{y}}^{2}}\end{array}$

Where it is implicit that all the functions are evaluated in t. Have I made an algebraic mistake, or is it not possible to find $||\dot{\overrightarrow{r}}(t)||$ by differentiation? - and if so, why?

asked 2022-07-16

I have two 2D unit vectors a and b. I'd like to find the rotation matrix that rotates a to b. The formulas I see online are for a rotation matrix are

$\left(\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)$

And I can get the angle between a and b with

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

My problem is that that doesn't give me the direction. For example, if $\theta $ is $\pi /2$ when maybe the matrix should use $-\pi /2$

$\left(\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)$

And I can get the angle between a and b with

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

My problem is that that doesn't give me the direction. For example, if $\theta $ is $\pi /2$ when maybe the matrix should use $-\pi /2$

asked 2022-08-14

Given vectors (1,3,5),(−2,−6,−10) and (2,6,10) determine whether the linear span of the above is a plane in ${\mathbb{R}}^{3}$

The vectors are linearly dependent nd hence do not form a basis and it is known that the set of linearly dependent vectors in R^2 are collinear.

So based on the above can it be said that linearly dependent vectors in R^3 will form a plane.

The vectors are linearly dependent nd hence do not form a basis and it is known that the set of linearly dependent vectors in R^2 are collinear.

So based on the above can it be said that linearly dependent vectors in R^3 will form a plane.