Point A(−4, 2, 5) and the two vectors, RAM = (20, 18, −10)? and RAN = (−10, 8, 15), define a triangle. Find (a) a unit vector perpendicular to the triangle;

(b) a unit vector in the plane of the triangle and perpendicular to RAN; (c) a unit vector in the plane of the triangle that bisects the interior angle at A. do only (a)

To find a unit vector perpendicular to the triangle defined by point \( A(-4, 2, 5) \) and vectors \( \mathbf{RAM} = (20, 18, -10) \) and \( \mathbf{RAN} = (-10, 8, 15) \), we first need to compute the cross product of the two vectors \( \mathbf{RAM} \) and \( \mathbf{RAN} \).

### Part (a): Unit Vector Perpendicular to the Triangle

The cross product of two vectors gives a vector that is perpendicular to the plane containing the two vectors. Given vectors:

\[
\mathbf{RAM} = (20, 18, -10), \quad \mathbf{RAN} = (-10, 8, 15)
\]

Let's compute the cross product \( \mathbf{RAM} \times \mathbf{RAN} \).

#### Step 1: Cross Product Calculation

The cross product of two vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \) is given by:

\[
\mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2, \, u_3v_1 - u_1v_3, \, u_1v_2 - u_2v_1)
\]

Substitute the components of \( \mathbf{RAM} \) and \( \mathbf{RAN} \):

\[
\mathbf{RAM} \times \mathbf{RAN} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
20 & 18 & -10 \\
-10 & 8 & 15
\end{vmatrix}
\]

Calculate each component:

\[
\mathbf{i}: \quad (18 \cdot 15) - (-10 \cdot 8) = 270 + 80 = 350
\]
\[
\mathbf{j}: \quad - \left( (20 \cdot 15) - (-10 \cdot (-10)) \right) = - (300 - 100) = -200
\]
\[
\mathbf{k}: \quad (20 \cdot 8) - (18 \cdot (-10)) = 160 + 180 = 340
\]

Thus, the cross product is:

\[
\mathbf{RAM} \times \mathbf{RAN} = (350, -200, 340)
\]

#### Step 2: Compute the Magnitude of the Cross Product

To find the unit vector, we first need the magnitude of the cross product:

\[
\| \mathbf{RAM} \times \mathbf{RAN} \| = \sqrt{350^2 + (-200)^2 + 340^2}
\]

Calculate the square of each component:

\[
350^2 = 122500, \quad (-200)^2 = 40000, \quad 340^2 = 115600
\]

\[
\| \mathbf{RAM} \times \mathbf{RAN} \| = \sqrt{122500 + 40000 + 115600} = \sqrt{278100}
\]

Compute the square root:

\[
\| \mathbf{RAM} \times \mathbf{RAN} \| = 527.33
\]

#### Step 3: Find the Unit Vector

Now, divide each component of the cross product by its magnitude to get the unit vector:

\[
\mathbf{n} = \frac{\mathbf{RAM} \times \mathbf{RAN}}{\| \mathbf{RAM} \times \mathbf{RAN} \|}
\]

\[
\mathbf{n} = \left( \frac{350}{527.33}, \frac{-200}{527.33}, \frac{340}{527.33} \right)
\]

Simplify each component:

\[
\mathbf{n} \approx \left( 0.664, -0.379, 0.645 \right)
\]

### Answer

The unit vector perpendicular to the triangle defined by point \(A\) and vectors \( \mathbf{RAM} \) and \( \mathbf{RAN} \) is:

\[
\boxed{\mathbf{n} \approx (0.664, -0.379, 0.645)}
\]

Would you like more details, or do you have any questions?

Here are some related questions you might find interesting:

1. How to compute the angle between two vectors using the dot product?
2. What is the geometric interpretation of the cross product?
3. How do you find the area of a triangle defined by three points in 3D space?
4. What are some applications of cross products in physics and engineering?
5. How would the result change if the vectors were linearly dependent?

**Tip:** Always double-check your vector arithmetic when performing cross products, as errors can propagate easily!

Learn how to find a unit vector perpendicular to a triangle defined by a point and two vectors in 3D space. This detailed solution involves computing the cross product of two vectors and finding its magnitude to derive the unit vector. Suitable for advanced high school students and anyone interested in 3D geometry.

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