Math Problem Statement

Solve a and b when a=3i-4j+2k and b = -i + 2j-3k

Solution

Given the vectors a=3i4j+2k\mathbf{a} = 3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k} and b=i+2j3k\mathbf{b} = -\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}, we can perform a variety of vector operations.

1. Magnitude of a\mathbf{a} and b\mathbf{b}

The magnitude (or length) of a vector v=xi+yj+zk\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} is given by:

v=x2+y2+z2|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}

For a=3i4j+2k\mathbf{a} = 3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}:

a=32+(4)2+22=9+16+4=29|\mathbf{a}| = \sqrt{3^2 + (-4)^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29}

For b=i+2j3k\mathbf{b} = -\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}:

b=(1)2+22+(3)2=1+4+9=14|\mathbf{b}| = \sqrt{(-1)^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

2. Dot Product ab\mathbf{a} \cdot \mathbf{b}

The dot product of two vectors a\mathbf{a} and b\mathbf{b} is calculated as:

ab=(3)(1)+(4)(2)+(2)(3)=386=17\mathbf{a} \cdot \mathbf{b} = (3)(-1) + (-4)(2) + (2)(-3) = -3 - 8 - 6 = -17

3. Cross Product a×b\mathbf{a} \times \mathbf{b}

The cross product a×b\mathbf{a} \times \mathbf{b} is a vector given by:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -4 & 2 \\ -1 & 2 & -3 \end{vmatrix}$$ This determinant expands as: $$\mathbf{a} \times \mathbf{b} = \mathbf{i}\left((-4)(-3) - (2)(2)\right) - \mathbf{j}\left((3)(-3) - (2)(-1)\right) + \mathbf{k}\left((3)(2) - (-4)(-1)\right)$$ Simplifying: $$\mathbf{a} \times \mathbf{b} = \mathbf{i}(12 - 4) - \mathbf{j}(-9 + 2) + \mathbf{k}(6 - 4)$$ $$\mathbf{a} \times \mathbf{b} = 8\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}$$ ### 4. Angle Between $$\mathbf{a}$$ and $$\mathbf{b}$$ The angle $$\theta$$ between two vectors is given by: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$$ Substituting the values: $$\cos \theta = \frac{-17}{\sqrt{29} \cdot \sqrt{14}}$$ $$\cos \theta = \frac{-17}{\sqrt{406}} \approx -0.844$$ Thus, the angle $$\theta$$ is: $$\theta \approx \cos^{-1}(-0.844) \approx 147.6^\circ$$ Would you like more details on any specific part or further calculations? --- **Here are some related questions:** 1. How do you interpret the cross product of two vectors geometrically? 2. What are the properties of the dot product and cross product of vectors? 3. How do you find the projection of one vector onto another? 4. Can you find a unit vector in the direction of $$\mathbf{a} \times \mathbf{b}$$? 5. What is the physical significance of the angle between two vectors? **Tip:** The cross product of two vectors results in a vector that is perpendicular to the plane containing the original vectors.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Cross Product
Vector Magnitude
Angle Between Vectors

Formulas

Magnitude of vector |v| = sqrt(x^2 + y^2 + z^2)
Dot product a · b = ax * bx + ay * by + az * bz
Cross product a × b = determinant of the matrix with unit vectors i, j, k and the components of a and b
Angle θ between vectors: cos θ = (a · b) / (|a| |b|)

Theorems

Pythagorean Theorem for vector magnitude
Dot Product and Cross Product Theorems

Suitable Grade Level

Grades 11-12