Question $p \cdot q=(2 i+j-k)(i-3 j+2 k)$

Studdy Solution

STEP 1

1. The vectors $2i + j - k$ and $i - 3j + 2k$ are expressed in the standard basis $\{i, j, k\}$ of $\mathbb{R}^3$ .
2. The dot product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ , where $\mathbf{a} = a_1 i + a_2 j + a_3 k$ and $\mathbf{b} = b_1 i + b_2 j + b_3 k$ .
3. The result of the dot product will be a scalar value.

STEP 2

1. Identify the components of the vectors $2i + j - k$ and $i - 3j + 2k$ .
2. Apply the dot product formula to these components.
3. Simplify the resulting expression to obtain the scalar value.

STEP 3

Identify the components of the vectors.
For the vector $2i + j - k$ , the components are: $a_1 = 2, \quad a_2 = 1, \quad a_3 = -1$
For the vector $i - 3j + 2k$ , the components are: $b_1 = 1, \quad b_2 = -3, \quad b_3 = 2$

STEP 4

Apply the dot product formula $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ to the identified components.
$(2i + j - k) \cdot (i - 3j + 2k) = (2)(1) + (1)(-3) + (-1)(2)$

STEP 5

Calculate each term in the dot product formula.
$(2)(1) = 2$ $(1)(-3) = -3$ $(-1)(2) = -2$

STEP 6

Add the calculated terms to obtain the final scalar value.
$2 + (-3) + (-2) = 2 - 3 - 2 = -3$
Solution: The dot product $p \cdot q$ is:
$(2i + j - k) \cdot (i - 3j + 2k) = -3$

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