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Chapter 14: Problem 17

Given three vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\), their triple vector product is defined to be \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c}\). For the vectors \(\boldsymbol{a}=4 i+2 \boldsymbol{j}+\boldsymbol{k}\) \(b=2 i-j+7 k\) and \(c=2 i-2 j+3 k\) verify that $$ (a \times b) \times c=(a \cdot c) b-(b \cdot c) a $$

Short Answer

Expert verified
Answer: No, the given identity is not verified for these particular vectors. We found that \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c} = -62i - 29j + 22k\) and \((\boldsymbol{a} \cdot \boldsymbol{c})\boldsymbol{b} - (\boldsymbol{b} \cdot \boldsymbol{c})\boldsymbol{a} = -94i - 61j + 22k\). These results are not equal.

Step by step solution

01

Compute the cross product \(\boldsymbol{a} \times \boldsymbol{b}\)

To compute the cross product \(\boldsymbol{a} \times \boldsymbol{b}\), we use the following formula: \[\begin{pmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} = i(a_2b_3 - a_3b_2) - j(a_1b_3 - a_3b_1) + k(a_1b_2 - a_2b_1)\] Substitute the components of \(\boldsymbol{a}\) and \(\boldsymbol{b}\): \[\boldsymbol{a} \times \boldsymbol{b} = i(2 \cdot 7 - 1 \cdot -1) - j(4 \cdot 7 - 1 \cdot 2) + k(4 \cdot -1 - 2 \cdot 2)\] \[\boldsymbol{a} \times \boldsymbol{b} = 14i + i - j(28 - 2) + k(-4 - 4)\] \[\boldsymbol{a} \times \boldsymbol{b} = 15i - 26j - 8k\]
02

Compute the triple vector product \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c}\)

Now, we need to compute the cross product \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c}\). Use the cross product formula again with \(\boldsymbol{a} \times \boldsymbol{b}\) and \(\boldsymbol{c}\): \[\boldsymbol{a \times b} \times \boldsymbol{c} = i([-26 \cdot 3 - (-8) \cdot (-2)]) - j(15 \cdot 3 - (-8) \cdot 2) + k(15 \cdot (-2) - (-26) \cdot 2)\] \[\boldsymbol{a \times b} \times \boldsymbol{c} = i(-78 + 16) - j(45 - 16) + k(-30 + 52)\] \[\boldsymbol{a \times b} \times \boldsymbol{c} = -62i - 29j + 22k\]
03

Compute the dot products \(\boldsymbol{a} \cdot \boldsymbol{c}\) and \(\boldsymbol{b} \cdot \boldsymbol{c}\)

Compute the dot products \(\boldsymbol{a} \cdot \boldsymbol{c}\) and \(\boldsymbol{b} \cdot \boldsymbol{c}\) using the following formula: \[a \cdot c = a_1c_1 + a_2c_2 + a_3c_3\] \[b \cdot c = b_1c_1 + b_2c_2 + b_3c_3\] For \(\boldsymbol{a} \cdot \boldsymbol{c}\): \[(4 \cdot 2) + (2 \cdot (-2)) + (1 \cdot 3) = 8 - 4 + 3 = 7\] For \(\boldsymbol{b} \cdot \boldsymbol{c}\): \[(2 \cdot 2) + (-1 \cdot (-2)) + (7 \cdot 3) = 4 + 2 + 21 = 27\]
04

Calculate \((\boldsymbol{a} \cdot \boldsymbol{c})\boldsymbol{b} - (\boldsymbol{b} \cdot \boldsymbol{c})\boldsymbol{a}\)

Now, using the values of \(\boldsymbol{a} \cdot \boldsymbol{c}\) and \(\boldsymbol{b} \cdot \boldsymbol{c}\), calculate \((\boldsymbol{a} \cdot \boldsymbol{c})\boldsymbol{b} - (\boldsymbol{b} \cdot \boldsymbol{c})\boldsymbol{a}\): \[((7)(2i - j + 7k) - (27)(4i + 2j + k))\] \[=(14i - 7j + 49k) - (108i + 54j + 27k)\] Collect like terms: =\[-94i - 61j + 22k\]
05

Compare the results from steps 2 and 4

We can now compare the results obtained in steps 2 and 4: - In step 2, we computed the triple vector product \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c}\) and found it equal to \(-62i - 29j + 22k\). - In step 4, we computed \((\boldsymbol{a} \cdot \boldsymbol{c})\boldsymbol{b} - (\boldsymbol{b} \cdot \boldsymbol{c})\boldsymbol{a}\), and found it equal to \(-94i - 61j + 22k\). As the two results are not equal, the given identity is not verified for these particular vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\), and \(\boldsymbol{c}\).

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Most popular questions from this chapter

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.

Line I has equation $$ \boldsymbol{r}_{1}=\left(\begin{array}{l} 2 \\ 3 \\ 5 \end{array}\right)+k\left(\begin{array}{l} 1 \\ 2 \\ 4 \end{array}\right) $$ Line II has equation $$ \boldsymbol{r}_{2}=\left(\begin{array}{c} -5 \\ 8 \\ 1 \end{array}\right)+l\left(\begin{array}{c} -6 \\ 7 \\ 0 \end{array}\right) $$ Different values of \(k\) give different points on line I. Similarly, different values of \(l\) give different points on line II. If the two lines intersect then \(\boldsymbol{r}_{1}=\boldsymbol{r}_{2}\) at the point of intersection. If you can find values of \(k\) and \(l\) which satisfy this condition then the two lines intersect. Show the lines intersect by finding these values and hence find the point of intersection.

Find the modulus of each of the following vectors: (a) \(\boldsymbol{r}=7 \boldsymbol{i}+3 \boldsymbol{j}\) (b) \(\boldsymbol{r}=17 \boldsymbol{i}\) (c) \(\boldsymbol{r}=2 \boldsymbol{i}-3 \boldsymbol{j}\) (d) \(r=-3 j\) (f) \(\boldsymbol{r}=a \boldsymbol{i}-b \boldsymbol{j}\)

Two unit vectors are parallel. What can you deduce about their scalar product?

Point P has coordinates \((7,8)\). Point \(Q\) has coordinates \((-2,4)\) (a) Draw a sketch showing \(\mathrm{P}\) and \(\mathrm{Q}\). (b) State the position vectors of \(\mathrm{P}\) and \(\mathrm{Q}\). (c) Find an expression for \(\overrightarrow{P Q}\). (d) Find \(|\overrightarrow{\mathrm{PQ}}|\).
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