Problem 1. Let u⃗ = (1, 2, 0), v⃗ = (0, 0, 3), and w⃗ = (0,...

Problem 1. Let $\vec{u} = (1, 2, 0)$, $\vec{v} = (0, 0, 3)$, and $\vec{w} = (0, 1, 0)$ be vectors in $\mathbb{R}^3$. Evaluate, if defined, the following expressions. If undefined, state so. a) $(\vec{u} \cdot \vec{v})\vec{w}$ b) $\vec{u} + \vec{w}$ c) $(\vec{u} \cdot \vec{w})/\vec{w}$ d) $(\vec{u} \cdot \vec{v})/(\vec{w} \cdot \vec{w})$

Transcript

00:01 We are given that u, v and w are vectors which belong to n -dimensional real space.

00:12 Okay, and we are given three statements.

00:15 The first statement is if the magnitude of vector u is equal to 4 and the magnitude of vector v is equal to 5 along with this condition that the magnitude of u plus v is equal to 7 then the dot product of u and v should be equal to 4.

00:44 Okay.

00:45 Now from the given information, let me just change it because this is the statement that we have to prove.

00:53 Okay.

00:54 However, let the given conditions be denoted by equation 1 which is the set of these equations.

01:02 Okay.

01:03 Now, the magnitude of u plus v square will be nothing but this u plus v dot u plus v which is equal to u plus v square.

01:26 By definition, this is so.

01:31 This is then equal to u plus v.

01:35 We can expand this like this.

01:41 U square plus v square plus 2 u dot v.

01:53 Okay.

01:55 Now, this is our equation.

02:00 Let this be equation 2.

02:02 Sorry, this is 2.

02:05 From equation 1 and equation 2, we can substitute the values of equation 1 and equation 2.

02:15 Right.

02:16 So doing that, u plus v whole square, it is given to us to be equal to 7 square.

02:23 Okay, and u square is equal to 4 square plus v square which is equal to 5 square.

02:35 These values have been substituted from equation 1.

02:38 Plus 2 u dot v.

02:42 Okay, so this implies that 2 u dot v is equal to 49 minus 16 minus 25 which is equal to 8.

03:00 So, this implies that u dot v is equal to 8 by 2 which is equal to 4.

03:11 This implies that u dot v is equal to 4.

03:17 This implies that statement 1 is true.

03:20 Now, for the second statement, the second statement states that suppose the magnitude of vector u is equal to 2, the magnitude of vector v is equal to 3.

03:36 Then, the magnitude of the difference between the vectors u and v will be less than 6.

03:45 So, taking a closer look at this equation, squaring both sides lhs and rhs, we get that u minus v whole square is less than 36.

04:00 Okay, so let's see if whether this is the case or not.

04:06 So as we did for the earlier statement, we can express u minus v whole square, the magnitude of u minus v whole square, it will be equal to u minus v into u minus b by a very definition.

04:26 So, this will be equal to u minus v whole square.

04:32 Right.

04:33 So, expanding this, this is u minus v sorry, u minus v whole square.

04:44 This is equal to, now we expand this, which is nothing but u square minus, plus v square, magnitude of v square, minus 2 u dot v.

05:03 Okay, so, suppose this be our set of equation 3 and this is our equation 4.

05:16 So, we substitute the values of equation 3 into equation 4.

05:21 From equation 3 and 4, we get u minus v whole square.

05:40 This will be then equal to, from equation 3, we get this is 2 square plus 3 square minus, now, this u dot v can also be expressed as the magnitude of vector u into the magnitude of vector v into cosine of the angle between them.

06:04 Theta is the angle between u and v.

06:11 Okay, so, now we arrive at this equation or condition...

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