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Algebra (all content)

Course: Algebra (all content)>Unit 19

Lesson 9: Adding vectors in magnitude & direction form

Adding vectors in magnitude & direction form (1 of 2)

Watch Sal add two vectors given in magnitude and direction form by breaking them down into components first. Created by Sal Khan.

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• At , he says the side is 3(sqrt3)/2. Why? Where did the sqrt of 3 come from?
• The sqrt(3) comes from the fact that this is a 30/60/90 triangle. You can search google for "30 60 90 triangles" and see more clearly (with pictures) why this is such.
• Wouldn't it be easier to just convert vectors a and b to their component forms and then add the components, getting a decimal approx. for the sum of a and b (call it c) and then use the magnitude formula to solve?

a = (3 cos 30, 3 sin 30)
b = (2 cos 135, 2 sin 135)
a + b = c = (3 cos 30 + 2 cos 135, 3 sin 30 + 2 sin 135)
c ~= (-1.53, -2,79)
|| c || = sqrt((-1.53)^2 + (-2.79)^2) ~= 3.18 (I know it's slightly off from the answer in the part 2)

I guess this video just took the component form of the vectors and changed it into exact representations of the trig functions?
• 𝒄 ≈ (1.18, 2.91)
You were in radian mode when you calculated the coordinates.

Nonetheless,
|𝒄| ≈ √((3 cos 30° + 2 cos 135°)² + (3 sin 30° + 2 sin 135°)²) ≈ 3.146
Let 𝜃 be the direction angle of 𝒄 ⇒ (since 𝒄 ∈ 𝐈) ⇒
⇒ 𝜃 = arctan((3 sin 30° + 2 sin 135°) ∕ (3 cos 30° + 2 cos 135°)) ≈ 67.89°

These are, of course, the same values that Sal got, so you're right, we don't need to worry about what the values of cos 30°, sin 30°, cos 135° and sin 135° are, as long as we don't care about the exact values of the magnitude and direction angle of 𝒄.

If we simplify the above expressions for |𝒄| and 𝜃, we get:
|𝒄| = √(13 + 3(√2 −√6))
𝜃 = arctan((9√3 + 6√2 + 6√6 + 8) ∕ 19)
Now, these expressions are hardly any easier to work with, but this doesn't mean that there aren't times when the simplified expressions actually are simpler, so it is still a good idea to at least try and simplify before taking the easy way out.
• But 3/2 and 3sqrt3/2 dont give 3
• Using the Pythagorean theorem, we know that
a^2 +b^2 =c^2
This means that the the length of the hypotenuse squared is equal to the sum of the two other sides squared.
If we find (3/2)^2 + (3sqrt3/2)^2, we get 9, which is the hypotenuse squared, therefore the actual length of the hypotenuse is 3.
• I was surprised by how fast the video went (at 2.11) to assuming we know about 30/60/90 triangles and their proportions. I tried to work this out for myself using my memory of sine (soh), cos (cah) and to check on these calculations tan (toa). But where do we learn these pre trigonometrical theses?
• It is a big time-saver to use the triangle facts (side ratios) for 30-60-90 triangles and 45-45-90 triangles. The 30-60-90 triangle is half of an equilateral triangle, while the 45-45-90 triangle is half of a square cut on the diagonal. It is worthwhile to review them because so many examples are based on these triangles and angles. In case you have forgotten, it is easy to use Pythagoras to rediscover. This is covered in 8th grade geometry and Geometry Basics and in the Geometry section (with practice exercises) in case you want a refresher.
Meanwhile:
30-60-90
short side = half of long side (hypotenuse or vector)
the third side is length of short side times √3
If vector is 60 units, short side is 30 units and third side is 30√3 Boom! Done just like that.
45-45-90
two sides are equal (it is half of a square)
the long side (hypotenuse or vector) will equal the side length times √2
If a vector is 12 units long, divide by √2 to get the side length, which gives 6√2
Example with Pythagoras: If the triangle is 30-60-90 and the vector is 13 units
a² + b² = c²
If the short side is a, we know it is 13/2 units (this could be either the x or the y, depending upon the orientation of the triangle.
(13/2)² + b² = 13²
b² = 13² - (13/2)²
b = √13² - (13/2)²
b = √169 - 169/4
b = √676/4 - 169/4
b = √507/4
b = ½√169 ∙ 3
b = ½13 √3
So, b = 13/2 √3

Since the short side was 13/2, the result for the third side is 13/2 `√3` which is exactly what is expected using the shortcut ratios.
Hope that helps.
• I don't know about others but in India we don't get calculator in exam, so how will I solve degree/radian problems...
• 2 pi radians equals 360 degrees. As a result, you can multiply your degrees by pi/180 to convert to radians. To go from radians to degrees, simply multiply by 180/pi
• what does sin stand for?
• That is the "sine function" from Trigonometry, typically used to describe the ratio of vertical motion to diagonal motion.
• Hi. How do you know that it is 30-60-90 triangle, not 45-45-90 ?
• Hello,
It will be a 30-60-90 angle triangle as one given angle is 30degree and other is a right angle. Applying the angle sum property of a triangle gives you the remaining angle, i.e. 60 degree.
• Wait...at , Sal says that the horizontal vector for the triangle (I hope I'm using that term correctly) is something multiplied by an imaginary number. However, I remember doing skills that had graphs featuring the y-axis as the imaginary axis. I don't really understand why he's doing that and I'm not sure if I missed something
• At how would we know that sine or cosine of 45 degrees is equal square-root of 2 over 2 (if Sal did not tell us) if we get decimal numbers on our calculators? How do we figure this out on our own for similar problems?
• Sines and cosines of certain degrees should be memorized. The cosine of 45 degrees in particular can be easily found to be √(2)/2 with just elementary geometry if you know the side ratios of a 45-45-90 triangle (which are 1 : 1 : √2).
(1 vote)
• Okay, given that vector 𝑎+ vector 𝑏=0, how is the magnitude of 𝑎 related to the magnitude of 𝑏? Using the example Sal used, I am still unable to answer this question.
(1 vote)
• When we add vectors, we arrange them tip-to-tail, then draw a new vector from 0 to the endpoint.

In our case, the new vector has length 0, so if we travel along a, b must then take us right back to the origin.

After we travel the length of a, we are a distance |a| from the origin (where |a| is the magnitude of a). We then travel another distance |b| and get back to the origin. So |a|=|b|. a and b have equal magnitudes.