Question

4) Three standard basis vectors are noted as p = 3i + 2k, q = 4i – 2; +3k and

r= -3i + 5j - 4k. Solve for :
i)
3p + 2q + 3r

ii)
10p + 74 + 8r​

Answer 1

i) 3p+2q+3r = 8i+11j

ii) 10p + 7q + 8r​= 34i+26j+9k

Explanation:

We are given vectors

p = 3i + 2k, q = 4i – 2j +3k and r= -3i + 5j - 4k

Note: There is typing mistake in q=4i-2;+3k there should be j instead of ; i.e 4i – 2j +3k

We need to find

i) 3p+2q+3r

Putting values of p, q and r

`3p+2q+3r\\=3(3i + 2k)+2(4i – 2j +3k)+3(-3i + 5j - 4k)\\=9i+6k+8i-4j+6k-9i+15j-12k\\=9i-9i+8i-4j+15j+6k+6k-12k\\=8i+11j+0k\\=8i+11j`

So, 3p+2q+3r = 8i+11j

ii) 10p + 7q + 8r​

Note: Considering 7q instead of 74

`10p + 7q + 8r​=10(3i + 2k)+7(4i – 2j+3k)+8(-3i + 5j - 4k)\\=30i+20k+28i-14j+21k-24i+40j-32k\\=30i+28i-24i-14j+40j+20k+21k-32k\\=34i+26j+9k`

So, 10p + 7q + 8r​= 34i+26j+9k