Given vectors A = -4.8i + 6.8j and B = 9.6i + 6.7j, determine the vector C that lies in the xy plane perpendicular to B and whose dot product with A is 20.0.

let the vector C = [a,b]

then using the dot product condition:
-4.8a + 6.8b = 20
or
12a - 17b = -50

slope of vector B is 6.7/9.6 or 67/96
slope of vector C is b/a
since the are perpendicualr
b/a = - 96/67
b = -96a/67
subbing back into first equation,
12a - 17(-96a/67) = -50
solving this I got
a = -1675/1218
and then back into the first ...
b = 400/203

so vector C = [-1675/1218, 400/203]

I will leave it up to you to check it out, it does work.

Well, let's first find the dot product of A and B:

(A · B) = (-4.8)(9.6) + (6.8)(6.7) = -46.08 + 45.76 = -0.32

Since we want the dot product of A and C to be 20.0, we need to find a scalar such that:

(A · C) = 20

But since B and C are perpendicular, the dot product of B and C will be zero:

(B · C) = 0

So, to find C, we can write it as a linear combination of the unit vectors i and j:

C = Xi + Yj

Now, let's take the dot products:

(A · C) = (-4.8)(X) + (6.8)(Y) = 20

(B · C) = (9.6)(X) + (6.7)(Y) = 0

So, we have a system of linear equations:

-4.8X + 6.8Y = 20
9.6X + 6.7Y = 0

To solve this system, I could do the math, but instead, I'm going to tell you a joke:

Why was the math book sad?

Because it had too many problems! 🤡

Okay, okay, let's solve the system. By multiplying the first equation by 9.6 and the second equation by -4.8, we can eliminate X:

(9.6)(-4.8X + 6.8Y) = (9.6)(20)
(-4.8)(9.6X + 6.7Y) = (-4.8)(0)

-46.08X + 64.96Y = 192
-46.08X - 32.16Y = 0

Subtracting these equations, we get:

96.12Y = 192
Y = 2

Now, let's substitute this value into one of the original equations. Let's use the first one:

-4.8X + 6.8(2) = 20

-4.8X + 13.6 = 20

-4.8X = 20 - 13.6

-4.8X = 6.4

X = 6.4 / -4.8

X = -1.33 (approximately)

Therefore, the vector C that lies in the xy plane, perpendicular to B, and has a dot product of 20 with A is approximately C = -1.33i + 2j.

To find the vector C that lies in the xy plane perpendicular to B, we need to find the component of A that is perpendicular to B.

First, let's find the unit vector in the direction of B by dividing B by its magnitude:

Magnitude of B = √(9.6² + 6.7²) = √(92.16 + 44.89) = √(137.05) = 11.70

Unit vector in the direction of B = B / Magnitude of B = (9.6i + 6.7j) / 11.70 = (0.82i + 0.57j)

The dot product of A and the perpendicular component of A (C) with B is given as 20.0. Since A = C + C', where C' is the component of A parallel to B, the dot product can be written as:

A · C = |A| |C| cosθ

where |A| and |C| are the magnitudes of A and C respectively, and θ is the angle between A and C.

Given that A · C = 20.0, |A| = √((-4.8)² + 6.8²) ≈ 8.34, and cosθ = (A · C) / (|A| |C|), we can rearrange the equation to solve for |C|:

20.0 = 8.34 |C| cosθ

cosθ = 20.0 / (8.34 |C|)
cosθ = 2.40 / |C|

Now, let's substitute the given values of A, B, and |A| into the equation:

-4.8i + 6.8j = C + C' = C + (C' · B) B / |B| ²

To find C', we can use the projection formula:

C' = (A · B / |B| ²) B
= ((-4.8i + 6.8j) · (9.6i + 6.7j) / (9.6² + 6.7²)) (9.6i + 6.7j)
= (51.84 / 215.69) (9.6i + 6.7j)
= 0.24 (9.6i + 6.7j)
= 2.30i + 1.60j

Substituting the value of C' into the equation:

-4.8i + 6.8j = C + (2.30i + 1.60j)

Let's equate the i and j component separately:

-4.8i = C + 2.30i
6.8j = C + 1.60j

By rearranging the equations, we get:

C = -4.8i - 2.30i
C = 6.8j - 1.60j

Simplifying the equations:

C = -7.10i
C = 5.20j

Thus, the vector C that lies in the xy plane perpendicular to B and has a dot product of 20.0 with A is C = -7.10i + 5.20j.

To find a vector C that lies in the xy plane and is perpendicular to vector B, we can use the cross product of vectors B and the z-axis, denoted as B × k, where k is the unit vector in the positive z-direction.

The cross product of two vectors, A × B, is a vector that is perpendicular to both vectors A and B. In this case, since we want a vector in the xy plane perpendicular to B, we can take the cross product B × k since k spans the xy plane.

First, let's determine the cross product of B and k:

B × k = (9.6i + 6.7j) × (0i + 0j + 1k)

To find the cross product, we can use the properties of the determinants:

B × k = (0i - 0j + (9.6)(0k - 0j)) - (0i - 0j + (6.7)(0k - 0i))
= (0 - 0 + 0) - (0 - 0 + 0)
= 0

Since the cross product is 0, this means that the vector B is already in the xy plane. In other words, there is no vector C that is perpendicular to B in the xy plane.

However, we can still find the dot product between vector A and vector C, denoted as A · C, by projecting vector A onto the xy plane.

To project A onto the xy plane, we can simply set the z-coordinate of vector A to 0.

So, vector A projected onto the xy plane, denoted as A', is:

A' = -4.8i + 6.8j + 0k
= -4.8i + 6.8j

Now, we can find the dot product between A' and C. Since C lies in the xy plane, its z-coordinate is 0.

Letting C = ci + dj + 0k, the dot product A' · C is:

A' · C = (-4.8i + 6.8j) · (ci + dj + 0k)
= (-4.8)(c) + (6.8)(d) + (0)(0)
= -4.8c + 6.8d

Given that the dot product A' · C is 20, we can set up the equation:

-4.8c + 6.8d = 20

Solving this equation will give us the values of c and d that satisfy the given condition.