Right Triangle ABC: Find 'm' & Projection Length

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Right Triangle ABC: Find 'm' & Projection Length

Hey guys! Today, we're diving into a cool problem involving triangles in 3D space. We've got triangle ABC with vertices A(4, 1, -3), B(m, 3, 5), and C(0, -3, 1). Our mission is twofold: first, we need to find the value of 'm' that makes this triangle a right triangle with the right angle at vertex A. Second, we'll calculate the length of the projection of side AB onto the hypotenuse AC. Buckle up, it's going to be a fun ride!

Part A: Finding the Value of 'm' for a Right Triangle at A

Okay, so the million-dollar question: how do we ensure that angle BAC is a right angle? The secret lies in the dot product! Remember, if two vectors are perpendicular, their dot product is zero. In our case, vectors AB and AC must be perpendicular for angle BAC to be 90 degrees.

Calculating Vectors AB and AC

First, let's find the vectors AB and AC. To do this, we subtract the coordinates of the initial point from the terminal point.

  • Vector AB = B - A = (m - 4, 3 - 1, 5 - (-3)) = (m - 4, 2, 8)
  • Vector AC = C - A = (0 - 4, -3 - 1, 1 - (-3)) = (-4, -4, 4)

Now that we have our vectors, let's move on to the dot product.

Applying the Dot Product

The dot product of AB and AC is calculated as follows:

AB · AC = (m - 4) * (-4) + (2) * (-4) + (8) * (4)

For the triangle to be right-angled at A, this dot product must equal zero:

(m - 4) * (-4) + (2) * (-4) + (8) * (4) = 0

Let's simplify this equation:

-4m + 16 - 8 + 32 = 0

-4m + 40 = 0

Now, solve for 'm':

-4m = -40

m = 10

So, there you have it! The value of 'm' that makes triangle ABC a right triangle at A is 10. This means that when m = 10, the vectors AB and AC are perpendicular, creating that crucial 90-degree angle.

Part B: Calculating the Projection of AB onto AC

Alright, now that we've nailed down the value of 'm', let's tackle the second part of our mission: finding the length of the projection of side AB onto the hypotenuse AC. In simpler terms, we want to know how much of AB lies along the direction of AC. This involves a bit of vector projection magic!

Understanding Vector Projection

The projection of vector AB onto vector AC, denoted as projACAB, is given by the formula:

projACAB = ((AB · AC) / ||AC||2) * AC

Where:

  • AB · AC is the dot product of AB and AC (which we already calculated).
  • ||AC|| is the magnitude (length) of vector AC.

Calculating the Magnitude of AC

First, we need to find the magnitude of vector AC. Recall that AC = (-4, -4, 4). The magnitude is calculated as:

||AC|| = √((-4)2 + (-4)2 + (4)2) = √(16 + 16 + 16) = √48 = 4√3

Therefore, ||AC||2 = (4√3)2 = 48.

Calculating the Projection Vector

We already know that AB · AC = 0 (since the triangle is right-angled at A when m = 10). However, let's pretend we didn't know that for demonstration purposes and use the original calculation:

AB = (10-4, 2, 8) = (6, 2, 8) AC = (-4, -4, 4) AB · AC = (6 * -4) + (2 * -4) + (8 * 4) = -24 - 8 + 32 = 0

So, the projection vector is:

projACAB = (0 / 48) * AC = 0 * (-4, -4, 4) = (0, 0, 0)

Finding the Length of the Projection

Since the projection vector is (0, 0, 0), the length of the projection is:

||projACAB|| = √(02 + 02 + 02) = 0

This makes sense because, in a right triangle, the projection of one leg onto the hypotenuse, from the right angle, is a point – it has zero length.

Alternative Approach and Potential Pitfalls

Now, let's consider a slightly different interpretation of the problem. Suppose the question meant the projection of AB onto BC (instead of AC), assuming AC is the hypotenuse. This is crucial because the initial interpretation leads to a trivial result (projection length of 0).

First, we need vector BC:

BC = C - B = (0 - 10, -3 - 3, 1 - 5) = (-10, -6, -4)

Now, let's calculate the dot product of AB and BC:

AB · BC = (6 * -10) + (2 * -6) + (8 * -4) = -60 - 12 - 32 = -104

Next, we find the magnitude squared of BC:

||BC||2 = (-10)2 + (-6)2 + (-4)2 = 100 + 36 + 16 = 152

Now, we calculate the projection vector:

projBCAB = (-104 / 152) * BC = (-104 / 152) * (-10, -6, -4)

projBCAB = (1040/152, 624/152, 416/152) = (65/19, 39/19, 26/19)

Finally, the length of this projection is:

||projBCAB|| = √((65/19)2 + (39/19)2 + (26/19)2)

||projBCAB|| = √(4225/361 + 1521/361 + 676/361) = √(6422/361) = √(2*3211)/19 = (√6422) / 19

This value is approximately 4.21. However, given the initial question, and the context of right triangle at A, projecting onto AC is more pertinent, yielding 0.

Conclusion

So, to wrap it up: for triangle ABC to be a right triangle at A, the value of 'm' must be 10. The length of the projection of AB onto AC is 0, given the right angle at A. If, alternatively, the question meant projection on BC, the process yields approximately 4.21. Always double-check the initial setup and question context to ensure accurate results! Keep up the great work, everyone!