Unite 3 0 1

How do you find a unit vector perpendicular to a 3-D plane formed by points (1,01),(0,2,2) and (3,3,0)?

1 Answer

Explanation:

Unite 3 lets you turn websites into apps on your Mac and change the way you use your computer forever. Creating your app. It all starts with the new Unite creation tool. Simply enter a name for your app, enter a URL, pick an icon, and launch your app. In addition to using a custom icon, Unite can automatically grab the HQ site favicon for you. Unity is the ultimate game development platform. Use Unity to build high-quality 3D and 2D games, deploy them across mobile, desktop, VR/AR, consoles or the Web, and connect with loyal and enthusiastic players and customers. Module 2 – Unit 3; Module 2 – Unit 3 You must first complete Module 2 – Unit 2 before viewing this Lesson. Please purchase the course before starting the lesson. This section discusses clients who have'Problems with Detoxification', as well as sample tests, and lab test analysis.

Our strategy will be to find two vectors in the plane, take their cross product to find a vector perpendicular to both of them (and thus to the plane), and then divide that vector by its measure to make it a unit vector.

Step 1) Find two vectors in the plane.

We will do this by finding the vector from #(1,0,1)# to #(0,2,2)# and from #(1,0,1)# to #(3,3,0)#. As all three points are in the plane, so will each of those vectors.

#vec(v_1) = (0,2,2)-(1,0,1) = (-1,2,1)#

#vec(v_2) = (3,3,0)-(1,0,1)=(2,3,-1)#

Honda wave 125 s manual. Step 2) Find a vector perpendicular to the plane.

If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of #vec(v_1)# and #vec(v_2)# to find a vector #vec(u)# perpendicular to the plane containing them.

#vec(u) = vec(v_1)xxvec(v_2)#

#= |(hat(i), hat(j), hat(k)), (-1, 2, 1), (2, 3, -1)|#

#=(2(-1)-1(3))hat(i)-((-1)(-1)-(1)(2))hat(j)+((-1)(3)-(2)(2))hat(k)#

#=-5hat(i)+hat(j)-7hat(k)#

#=(-5, 1, -7)#

Step 3) Turn #vec(u)# into a unit vector.

A unit vector is a vector whose measure is #1#. Using the fact that for any vector #vec(v)# and scalar #c#, we have #||cvec(v)|| = c||vec(v)||#, we will find #||vec(u)|| = u#, then divide by #u#.

#||vec(u)/u|| = ||vec(u)||/u = u/u = 1#

As multiplying by a scalar does not change the direction of a vector, this will be a unit vector perpendicular to the plane. Proceeding,

#||vec(u)|| = sqrt((-5)^2+1^2+(-7)^2) = sqrt(75)=5sqrt(3)# https://herezfile202.weebly.com/mac-video-downloader-1-0-intelk-download-free.html.

Thus, our final result is

#vec(u)/u = ((-5','1','-7))/(5sqrt(3)) = (-1/sqrt(3), 1/(5sqrt(3)), -7/(5sqrt(3)))#

Related questions

Show Mobile NoticeShow All NotesHide All Notes

Section 1-8 : Tangent, Normal and Binormal Vectors

In this section we want to look at an application of derivatives for vector functions. Actually, there are a couple of applications, but they all come back to needing the first one.

In the past we've used the fact that the derivative of a function was the slope of the tangent line. With vector functions we get exactly the same result, with one exception.

Given the vector function, (vec rleft( t right)), we call (vec r'left( t right)) the tangent vector provided it exists and provided (vec r'left( t right) ne vec 0). The tangent line to (vec rleft( t right)) at (P) is then the line that passes through the point (P) and is parallel to the tangent vector, (vec r'left( t right)). Note that we really do need to require (vec r'left( t right) ne vec 0) in order to have a tangent vector. If we had [vec r'left( t right) = vec 0]we would have a vector that had no magnitude and so couldn't give us the direction of the tangent.

Also, provided (vec r'left( t right) ne vec 0), the unit tangent vector to the curve is given by,

While, the components of the unit tangent vector can be somewhat messy on occasion there are times when we will need to use the unit tangent vector instead of the tangent vector.

Example 1 Find the general formula for the tangent vector and unit tangent vector to the curve given by (vec rleft( t right) = {t^2},vec i + 2sin t,vec j + 2cos t,vec k). Show Solution

First, by general formula we mean that we won't be plugging in a specific (t) and so we will be finding a formula that we can use at a later date if we'd like to find the tangent at any point on the curve. Mail pilot 3 39 epizoda. With that said there really isn't all that much to do at this point other than to do the work.

Here is the tangent vector to the curve.

[vec r'left( t right) = 2t,vec i + 2cos t,vec j - 2sin t,vec k]

To get the unit tangent vector we need the length of the tangent vector.

[begin{align*}left| {vec r'left( t right)} right| & = sqrt {4{t^2} + 4{{cos }^2}t + 4{{sin }^2}t} & = sqrt {4{t^2} + 4} end{align*}]

The unit tangent vector is then,

[begin{align*}vec Tleft( t right) & = frac{1}{{sqrt {4{t^2} + 4} }}left( {2t,vec i + 2cos ,t,vec j - 2sin t,vec k} right) & = frac{{2t}}{{sqrt {4{t^2} + 4} }},vec i + frac{{2cos t}}{{sqrt {4{t^2} + 4} }}vec j - frac{{2sin t}}{{sqrt {4{t^2} + 4} }},vec kend{align*}] Example 2 Find the vector equation of the tangent line to the curve given by (vec rleft( t right) = {t^2},vec i + 2sin t,vec j + 2cos t,vec k) at (displaystyle t = frac{pi }{3}). Show Solution

First, we need the tangent vector and since this is the function we were working with in the previous example we can just reuse the tangent vector from that example and plug in (t = frac{pi }{3}).

[vec r'left( {frac{pi }{3}} right) = frac{{2pi }}{3},vec i + 2cos left( {frac{pi }{3}} right),vec j - 2sin left( {frac{pi }{3}} right),vec k = frac{{2pi }}{3},vec i + vec j - sqrt 3 ,vec k]

We'll also need the point on the line at (t = frac{pi }{3}) so,

[vec rleft( {frac{pi }{3}} right) = frac{{{pi ^2}}}{9},vec i + sqrt 3 ,vec j + ,vec k]

The vector equation of the line is then, Microsoft office 2008.

[vec rleft( t right) = leftlangle {frac{{{pi ^2}}}{9},sqrt 3 ,1} rightrangle + tleftlangle {frac{{2pi }}{3},1, - sqrt 3 } rightrangle ]

Before moving on let's note a couple of things about the previous example. First, we could have used the unit tangent vector had we wanted to for the parallel vector. Routebuddy v4 0. However, that would have made for a more complicated equation for the tangent line.

Second, notice that we used (vec rleft( t right)) to represent the tangent line despite the fact that we used that as well for the function. Do not get excited about that. The (vec rleft( t right)) here is much like (y) is with normal functions. With normal functions, (y) is the generic letter that we used to represent functions and (vec rleft( t right)) tends to be used in the same way with vector functions.

Next, we need to talk about the unit normal and the binormal vectors.

The unit normal vector is defined to be,

The unit normal is orthogonal (or normal, or perpendicular) to the unit tangent vector and hence to the curve as well. We've already seen normal vectors when we were dealing with Equations of Planes. They will show up with some regularity in several Calculus III topics.

The definition of the unit normal vector always seems a little mysterious when you first see it. It follows directly from the following fact.

Fact

Suppose that (vec rleft( t right)) is a vector such that (left| {vec rleft( t right)} right| = c) for all (t). Then (vec r'left( t right)) is orthogonal to (vec rleft( t right)).

To prove this fact is pretty simple. From the fact statement and the relationship between the magnitude of a vector and the dot product we have the following.

Unite 3 0 1 =

[vec rleft( t right),centerdot ,vec rleft( t right) = {left| {vec rleft( t right)} right|^2} = {c^2},{mbox{for all }}t]

Now, because this is true for all (t) we can see that,

[frac{d}{{dt}}left( {vec rleft( t right),centerdot ,vec rleft( t right)} right) = frac{d}{{dt}}left( {{c^2}} right) = 0]

Also, recalling the fact from the previous section about differentiating a dot product we see that,

[frac{d}{{dt}}left( {vec rleft( t right),centerdot ,vec rleft( t right)} right) = vec r'left( t right),centerdot ,vec rleft( t right) + vec rleft( t right),centerdot ,vec r'left( t right) = 2vec r'left( t right),centerdot ,vec rleft( t right)]

Or, upon putting all this together we get,

[2vec r'left( t right),centerdot ,vec rleft( t right) = 0hspace{0.25in}hspace{0.25in} Rightarrow hspace{0.25in}hspace{0.25in}vec r'left( t right),centerdot ,vec rleft( t right) = 0]

Therefore (vec r'left( t right)) is orthogonal to (vec rleft( t right)).

The definition of the unit normal then falls directly from this. Because (vec Tleft( t right)) is a unit vector we know that (left| {vec Tleft( t right)} right| = 1) for all (t) and hence by the Fact (vec T'left( t right)) is orthogonal to (vec Tleft( t right)). However, because (vec Tleft( t right)) is tangent to the curve, (vec T'left( t right)) must be orthogonal, or normal, to the curve as well and so be a normal vector for the curve. All we need to do then is divide by (left| {vec T'left( t right)} right|) to arrive at a unit normal vector.

Next, is the binormal vector. The binormal vector is defined to be,

Because the binormal vector is defined to be the cross product of the unit tangent and unit normal vector we then know that the binormal vector is orthogonal to both the tangent vector and the normal vector.

Example 3 Find the normal and binormal vectors for (vec rleft( t right) = leftlangle {t,3sin t,3cos t} rightrangle ). Show Solution

We first need the unit tangent vector so first get the tangent vector and its magnitude.

[begin{align*}vec r'left( t right) & = leftlangle {1,3cos t, - 3sin t} rightrangle left| {vec r'left( t right)} right| & = sqrt {1 + 9{{cos }^2}t + 9{{sin }^2}t} = sqrt {10} end{align*}]

The unit tangent vector is then,

[vec Tleft( t right) = leftlangle {frac{1}{{sqrt {10} }},frac{3}{{sqrt {10} }}cos t, - frac{3}{{sqrt {10} }}sin t} rightrangle ]

The unit normal vector will now require the derivative of the unit tangent and its magnitude.

[begin{align*}vec T'left( t right) & = leftlangle {0, - frac{3}{{sqrt {10} }}sin t, - frac{3}{{sqrt {10} }}cos t} rightrangle left| {vec T'left( t right)} right| & = sqrt {frac{9}{{10}}{{sin }^2}t + frac{9}{{10}}{{cos }^2}t} = sqrt {frac{9}{{10}}} = frac{3}{{sqrt {10} }}end{align*}]

The unit normal vector is then,

United 301 Flight Status

[2vec r'left( t right),centerdot ,vec rleft( t right) = 0hspace{0.25in}hspace{0.25in} Rightarrow hspace{0.25in}hspace{0.25in}vec r'left( t right),centerdot ,vec rleft( t right) = 0]

Therefore (vec r'left( t right)) is orthogonal to (vec rleft( t right)).

The definition of the unit normal then falls directly from this. Because (vec Tleft( t right)) is a unit vector we know that (left| {vec Tleft( t right)} right| = 1) for all (t) and hence by the Fact (vec T'left( t right)) is orthogonal to (vec Tleft( t right)). However, because (vec Tleft( t right)) is tangent to the curve, (vec T'left( t right)) must be orthogonal, or normal, to the curve as well and so be a normal vector for the curve. All we need to do then is divide by (left| {vec T'left( t right)} right|) to arrive at a unit normal vector.

Next, is the binormal vector. The binormal vector is defined to be,

Because the binormal vector is defined to be the cross product of the unit tangent and unit normal vector we then know that the binormal vector is orthogonal to both the tangent vector and the normal vector.

Example 3 Find the normal and binormal vectors for (vec rleft( t right) = leftlangle {t,3sin t,3cos t} rightrangle ). Show Solution

We first need the unit tangent vector so first get the tangent vector and its magnitude.

[begin{align*}vec r'left( t right) & = leftlangle {1,3cos t, - 3sin t} rightrangle left| {vec r'left( t right)} right| & = sqrt {1 + 9{{cos }^2}t + 9{{sin }^2}t} = sqrt {10} end{align*}]

The unit tangent vector is then,

[vec Tleft( t right) = leftlangle {frac{1}{{sqrt {10} }},frac{3}{{sqrt {10} }}cos t, - frac{3}{{sqrt {10} }}sin t} rightrangle ]

The unit normal vector will now require the derivative of the unit tangent and its magnitude.

[begin{align*}vec T'left( t right) & = leftlangle {0, - frac{3}{{sqrt {10} }}sin t, - frac{3}{{sqrt {10} }}cos t} rightrangle left| {vec T'left( t right)} right| & = sqrt {frac{9}{{10}}{{sin }^2}t + frac{9}{{10}}{{cos }^2}t} = sqrt {frac{9}{{10}}} = frac{3}{{sqrt {10} }}end{align*}]

The unit normal vector is then,

United 301 Flight Status

Unite 3 0 1 Sezonas

[vec Nleft( t right) = frac{{sqrt {10} }}{3}leftlangle {0, - frac{3}{{sqrt {10} }}sin t, - frac{3}{{sqrt {10} }}cos t} rightrangle = leftlangle {0, - sin t, - cos t} rightrangle ]

United 3015

Finally, the binormal vector is,

United 3001

[begin{align*}vec Bleft( t right) & = vec Tleft( t right) times vec Nleft( t right) & = left| {begin{array}{*{20}{c}}{vec i}&{vec j}&{vec k}{frac{1}{{sqrt {10} }}}&{frac{3}{{sqrt {10} }}cos t}&{ - frac{3}{{sqrt {10} }}sin t}0&{ - sin t}&{ - cos t}end{array}} right|,begin{array}{*{20}{c}}{vec i}&{vec j}{frac{1}{{sqrt {10} }}}&{frac{3}{{sqrt {10} }}cos t}0&{ - sin t}end{array} & = - frac{3}{{sqrt {10} }}{cos ^2}t,vec i - frac{1}{{sqrt {10} }}sin t,vec k + frac{1}{{sqrt {10} }}cos t,vec j - frac{3}{{sqrt {10} }}{sin ^2}t,vec i & = - frac{3}{{sqrt {10} }},vec i + frac{1}{{sqrt {10} }}cos t,vec j - frac{1}{{sqrt {10} }}sin t,vec kend{align*}]