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Chapter 16: Problem 8

For \(\boldsymbol{a}=(1,2,3), \boldsymbol{b}=(-2,3,-4), \boldsymbol{c}=(0,4,-1)\), find $$ 3 a+2 b-3 c $$

Short Answer

Expert verified
The resulting vector is \((-1, 0, 4)\).

Step by step solution

01

Understand the Expression

The given expression is a vector operation involving three vectors \( \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \) and involves scalar multiplication and vector addition/subtraction. We need to compute \( 3\boldsymbol{a} + 2\boldsymbol{b} - 3\boldsymbol{c} \).
02

Compute 3a

Multiply each component of the vector \( \boldsymbol{a} = (1, 2, 3) \) by 3. This yields: \[ 3\boldsymbol{a} = (3 \times 1, 3 \times 2, 3 \times 3) = (3, 6, 9) \]
03

Compute 2b

Multiply each component of the vector \( \boldsymbol{b} = (-2, 3, -4) \) by 2. This yields: \[ 2\boldsymbol{b} = (2 \times -2, 2 \times 3, 2 \times -4) = (-4, 6, -8) \]
04

Compute -3c

Multiply each component of the vector \( \boldsymbol{c} = (0, 4, -1) \) by -3. This yields: \[ -3\boldsymbol{c} = (-3 \times 0, -3 \times 4, -3 \times -1) = (0, -12, 3) \]
05

Combine the Results

Now, add the results of the previous steps together: \( 3\boldsymbol{a} + 2\boldsymbol{b} - 3\boldsymbol{c} = (3, 6, 9) + (-4, 6, -8) + (0, -12, 3) \).Compute each component separately:First component: \( 3 - 4 + 0 = -1 \)Second component: \( 6 + 6 - 12 = 0 \)Third component: \( 9 - 8 + 3 = 4 \)Thus, the resulting vector is \((-1, 0, 4)\).

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Key Concepts

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

Scalar Multiplication
Scalar multiplication is an operation where each component of a vector is multiplied by a scalar (a real number). This process scales the vector, which means it makes the vector longer or shorter, depending on the value of the scalar.

To perform scalar multiplication:
  • Take each component of the vector.
  • Multiply each by the scalar.
  • Reassemble the results into a new vector.
For example, if you have vector \( \boldsymbol{a} = (1, 2, 3) \) and want to multiply it by a scalar 3, you compute:
\[ 3 \boldsymbol{a} = (3 \times 1, 3 \times 2, 3 \times 3) = (3, 6, 9) \]
In this way, each component of \( \boldsymbol{a} \) is scaled by 3. This operation maintains the direction of the original vector, unless scaled by a negative, which reverses the direction.
Vector Addition
Vector addition involves combining the corresponding components of two vectors to form a new vector. This operation essentially "adds" the vectors together.
  • Start with two vectors, say \( \boldsymbol{u} = (u_1, u_2, u_3) \) and \( \boldsymbol{v} = (v_1, v_2, v_3) \).
  • Add the corresponding components: \( (u_1 + v_1, u_2 + v_2, u_3 + v_3) \).
For instance, if you have \( \boldsymbol{a} = (3, 6, 9) \) and \( \boldsymbol{b} = (-4, 6, -8) \), the vector addition is conducted as follows:
\[ \boldsymbol{a} + \boldsymbol{b} = (3 + (-4), 6 + 6, 9 + (-8)) = (-1, 12, 1) \]
This result represents the combination of the two vector effects in a single vector outcome.
Vector Subtraction
Vector subtraction works similarly to vector addition, but instead of adding, you subtract the corresponding components of the vectors.

Here鈥檚 how to perform vector subtraction:
  • Given two vectors \( \boldsymbol{u} = (u_1, u_2, u_3) \) and \( \boldsymbol{v} = (v_1, v_2, v_3) \).
  • Subtract the components: \( (u_1 - v_1, u_2 - v_2, u_3 - v_3) \).
An example would be with vectors \( 2\boldsymbol{b} = (-4, 6, -8) \) and \(-3\boldsymbol{c} = (0, -12, 3) \). Subtraction happens as follows:
\[ \boldsymbol{b} - \boldsymbol{c} = (-4 - 0, 6 - (-12), -8 - 3) = (-4, 18, -11) \]
This operation effectively finds the vector pointing from one to the other.

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Most popular questions from this chapter

The quantity \(\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})\) is called a triple scalar product. Show that (i) \(a \cdot(b \times c)=c \cdot(a \times b)=b \cdot(c \times a)\) (ii) \(\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})=\left|\begin{array}{ccc}a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z}\end{array}\right|\) (determinant)

For \(\boldsymbol{a}=(1,3,-2), \boldsymbol{b}=(0,3,1), \boldsymbol{c}=(0,-1,2)\), find:$$ a \times(b \times c),(a \times b) \times c $$

For \(\boldsymbol{a}=(1,3,-2), \boldsymbol{b}=(0,3,1), \boldsymbol{c}=(0,-1,2)\), find:$$ (a \times c) \cdot b $$

Differentiate with respect to \(t\). \(2 t \boldsymbol{i}+3 t^{2} \boldsymbol{j}\)

Two sides of the triangle \(\mathrm{ABC}\) are \(\overrightarrow{\mathrm{AB}}=(2,1,-1)\) and \(\overrightarrow{\mathrm{AC}}=(3,2,0) .\) Find \(\overrightarrow{\mathrm{BC}}\) Find (i) the vector \(\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right)\) with the given initial point \(\mathrm{P}\left(x_{1}, y_{1}, z_{1}\right)\) and terminal point \(\mathrm{Q}\left(x_{2}, y_{2}, z_{2}\right)\), (ii) the length of \(\boldsymbol{a}\), (iii) the unit vector parallel to \(\boldsymbol{a}\).
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