Hello,

Can somebody help me with the following two problems?

1. Find a value of a such that the vector <a, 1,1> makes an angle of 45 degrees with the vector <1,2,1> or show that no such a exists.

---I started by using theta = arccos (u*v)/(absu*absv) must equal (sqrt2)/2 and attempting to solve from there...but I'm not sure if that's the best way to proceed.

2. Let u = 3i + j; v = 5i - 2j; and w = i - j. Find scalars, a and b such that u = av +bw.

I have no idea how to proceed with this one.

I would greatly appreciate any help!!!

u = <a, 1,1>

v = <1,2,1>

u dot v = 1/2 sqrt(2) |u| |v| ---->

(u dot v)^2 = 1/2 |u|^2 |v|^2 ---->

(a+3)^2 = 1/2 (a^2 + 2)*6 ---->

2 a^2 -6 a -3 = 0

Let u = 3i + j; v = 5i - 2j; and w = i - j. Find scalars, a and b such that u = av +bw.

av+bw =

(5a +b)i + (-2a-b)j

av+bw = u --->

5a + b = 3

2a + b = -1

Sure, I can help you with both of these problems.

1. To find a value of a such that the vector <a, 1, 1> makes an angle of 45 degrees with the vector <1, 2, 1>, you're on the right track using the formula theta = arccos((u*v)/(|u|*|v|)). Let's break it down step by step:

- First, let's find the dot product of the two vectors:
u * v = a*1 + 1*2 + 1*1 = a + 2 + 1 = a + 3

- Next, let's find the magnitudes of the two vectors:
|u| = sqrt(a^2 + 1^2 + 1^2) = sqrt(a^2 + 2)
|v| = sqrt(1^2 + 2^2 + 1^2) = sqrt(6)

- Now, substituting these values into the equation, we have:
theta = arccos((a + 3)/(sqrt(a^2 + 2) * sqrt(6)))

- We know that theta should be 45 degrees, so we have:
arccos((a + 3)/(sqrt(a^2 + 2) * sqrt(6))) = 45 degrees

- To solve for a, let's start by taking the cosine of both sides:
cos(arccos((a + 3)/(sqrt(a^2 + 2) * sqrt(6)))) = cos(45 degrees)

- The cosine and arccos cancel out, leaving us with:
(a + 3)/(sqrt(a^2 + 2) * sqrt(6)) = cos(45 degrees)

- Simplifying the right side, we have:
(a + 3)/(sqrt(a^2 + 2) * sqrt(6)) = sqrt(2)/2

- Cross-multiplying to get rid of the fractions:
2(a + 3) = sqrt(2) * sqrt(a^2 + 2) * sqrt(6)

- Squaring both sides to get rid of the square roots:
4(a + 3)^2 = 2(a^2 + 2) * 6

- Expanding and simplifying:
4(a^2 + 6a + 9) = 12a^2 + 24

- Rearranging and simplifying:
4a^2 + 24a + 36 = 12a^2 + 24

- Bringing all terms to one side to form a quadratic equation:
8a^2 - 24a + 12 = 0

- Dividing through by 4:
2a^2 - 6a + 3 = 0

Now you can use the quadratic formula or factoring to solve for a. I'll let you take it from here.

2. To find scalars a and b such that u = av + bw, we need to express u as a linear combination of v and w. Recall that u is given as 3i + j, v is given as 5i - 2j, and w is given as i - j.

- We want to find scalars a and b such that:
3i + j = av + bw

- Expanding the right side using the given values of v and w:
3i + j = a(5i - 2j) + b(i - j)

- Expanding further:
3i + j = (5ai - 2aj) + (bi - bj)

- Matching the coefficients of i and j on both sides, we have the following equations:
3 = 5a + b
1 = -2a - b

- We now have a system of two equations in two variables (a and b). You can solve this system using any suitable method, such as substitution or elimination.

I'll leave it up to you to solve this system and find the values of a and b. Let me know if you need any further assistance.