Q. 25

Question

In Exercises 22–29 compute the indicated quantities when u=(2,1,3), v=(4,0,1), and w=(2,6,5)

(u×v)×w and u×(v×w)

Step-by-Step Solution

Verified
Answer

The value of (u×v)×w=94i-3j+34k and u×(v×w)=90i-30j+50k.

1Step 1. Given Information

In Exercises 22–29 compute the indicated quantities whenu=(2,1,3), v=(4,0,1), and w=(2,6,5)

We have to find the value of (u×v)×w and u×(v×w)

2Step 2. Firstly finding the value of ( u × v ) × w

The cross product of u×v=detijk21-3401

Now solving the cross product.

u×v=((1)(1)(3)(0))i+((2)(1)(-3)(4))j+((2)(0)(1)(4))ku×v=(1-0)i+(2+12)j+(04)ku×v=1i+14j-4k

3Step 3. Now the vectors of u × v   is   ( 1 , 14 , - 4 ) .

Now finding the value of (u×v)×w

(u×v)×w=detijk114-4-265(u×v)×w=((14)(5)(4)(6))i+((1)(5)(-4)(-2))j+((1)(6)(14)(-2))k(u×v)×w=(70+24)i+(5-8)j+(6+28)k(u×v)×w=94i-3j+34k

4Step 4. Firstly finding the value of v × w

The cross product of v×w=detijk401-265

v×w=((0)(5)(1)(6))i+((4)(5)(1)(-2))j+((4)(6)(0)(-2))kv×w=(0-6)i+(20+2)j+(24+0)kv×w=-6i+22j+24k

5Step 5. Now the vectors of v × w   is   ( - 6 , 22 , 24 )

u×(v×w)=detijk21-3-62224u×(v×w)=((1)(24)(3)(22))i+((2)(24)(-3)(-6))j+((2)(22)(1)(-6))ku×(v×w)=(24+66)i+(48-18)j+(44+6)ku×(v×w)=90i-30j+50k