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Chapter 2: Problem 153

Determine the measure of angle \(A\) in triangle \(A B C,\) where \(A(1,1,8), \quad B(4,-3,-4),\) and \(C(-3,1,5)\) . Express your answer in degrees rounded to two decimal places.

Short Answer

Expert verified
Angle A measures approximately 67.98 degrees.

Step by step solution

01

Calculate Vector AB

Use the coordinates of points \( A \) and \( B \) to find the vector \( \overrightarrow{AB} \). Subtract the coordinates of \( A \) from \( B \): \[ \overrightarrow{AB} = B - A = (4 - 1, -3 - 1, -4 - 8) = (3, -4, -12) \]
02

Calculate Vector AC

Use the coordinates of points \( A \) and \( C \) to find the vector \( \overrightarrow{AC} \). Subtract the coordinates of \( A \) from \( C \): \[ \overrightarrow{AC} = C - A = (-3 - 1, 1 - 1, 5 - 8) = (-4, 0, -3) \]
03

Calculate Dot Product of Vectors AB and AC

Calculate the dot product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (3)(-4) + (-4)(0) + (-12)(-3) = -12 + 0 + 36 = 24 \]
04

Calculate Magnitude of Vectors

Find the magnitude of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \). For \( \overrightarrow{AB} \): \[ \| \overrightarrow{AB} \| = \sqrt{3^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \] For \( \overrightarrow{AC} \): \[ \| \overrightarrow{AC} \| = \sqrt{(-4)^2 + 0^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \]
05

Calculate Cosine of Angle A

Use the dot product and magnitudes to find \( \cos A \) using the formula: \[ \cos A = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\|\overrightarrow{AB}\| \cdot \|\overrightarrow{AC}\|} = \frac{24}{13 \times 5} = \frac{24}{65} \]
06

Calculate Angle A

Use the inverse cosine function to find \( \angle A \). Convert the result from radians to degrees: \[ A = \cos^{-1}\left(\frac{24}{65}\right) \approx 67.98^{\circ} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a way to multiply two vectors that yields a scalar value. You calculate it by multiplying the corresponding components of each vector and then summing those products. For example, if you have two vectors, \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is computed as:
  • \( u_1 \times v_1 + u_2 \times v_2 + u_3 \times v_3 \)
This calculation helps determine the angle between the vectors. A crucial concept here is that the dot product is zero if the vectors are perpendicular to each other.
In our exercise, we use the dot product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) to find the cosine of the angle at point A. The result of the dot product was found to be 24, which plays a pivotal role in determining the angle itself.
Vector Magnitude
Vector magnitude refers to the length or size of a vector. To compute this, you take the square root of the sum of the squares of its components. For any vector \( \mathbf{v} = (v_1, v_2, v_3) \), its magnitude \( \| \mathbf{v} \| \) is given by:
  • \( \sqrt{v_1^2 + v_2^2 + v_3^2} \)
This calculation is important when you want to normalize vectors or find distances.
In the exercise, we calculated the magnitudes of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) to be 13 and 5, respectively. These magnitudes are essential components when using the cosine law in vectors to find the angle.
Cosine Law in Vectors
The cosine law in vectors helps find the angle between two vectors using the dot product and their magnitudes. This law is very similar to the cosine rule in trigonometry, which relates the lengths of sides of a triangle to the cosine of one of its angles. The formula for finding the cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is:
  • \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \cdot \|\mathbf{v}\|} \)
This equation shows how the dot product and magnitudes are combined to capture the directional relationship between the vectors.
In our problem, the calculation of \( \cos A \) was achieved by substituting the dot product (24) and the magnitudes (13 and 5) into the formula. The result, \( \frac{24}{65} \), gives us the cosine of the angle, which we then convert into degrees to find the measure of angle A, approximately 67.98 degrees.

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Most popular questions from this chapter

[T]Let \(\mathbf{r}(t)=\left\langle t, 2 t^{2}, 4 t^{2}\right\rangle\) be the position vector of a particle at time \(t\) (in seconds), where \(t \in[0,10]\) (here the components of \(\mathbf{r}\) are expressed in centimeters). a. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places. b. Use a CAS to visualize the path of the particle defined by the points \(\left(t, 2 t^{2}, 4 t^{2}\right), \quad\) where \(t \in[0,60] .\)

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates. $$x^{2}+y^{2}+z^{2}=144$$

Consider \(\mathbf{r}(t)=\langle\cos t, \sin t, 2 t\rangle\) the position vector of a particle at time \(t \in[0,30],\) where the components of \(\mathbf{r}\) are expressed in centimeters and time in seconds. Let \(\overrightarrow{O P}\) be the position vector of the particle after 1 sec. a. Determine unit vector \(\mathbf{B}(t)\) (called the bi normal unit vector) that has the direction of cross product vector \(\mathbf{v}(t) \times \mathbf{a}(t),\) where \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\) are the instantaneous velocity vector and, respectively, the acceleration vector of the particle after \(t\) seconds. b. Use a CAS to visualize vectors \(\mathbf{v}(1), \mathbf{a}(1),\) and \(\mathbf{B}(1)\) as vectors statting at point \(P\) along with the path of the particle.

Determine a unit vector perpendicular to the plane passing through the \(z\) -axis and point \(A(3,1,-2)\) .

Consider line \(L\) of symmetric equations \(x-2=-y=\frac{z}{2}\) and point \(A(1,1,1) .\) a. Find parametric equations for a line parallel to L that passes through point A. b. Find symmetric equations of a line skew to L and that passes through point A. c. Find symmetric equations of a line that intersects L and passes through point A.
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