Question

The dot products of a vector with the vectors $(\hat{i}+\hat{j}-3\hat{k}),(\hat{i}+3\hat{j}-2\hat{k})$ and $(2\hat{i}+\hat{j}+4\hat{k})$ are $0,5,8$ respectively. Find the vector.

Answer 1

Let the vector be $ai+bj+ck$

$(ai+bj+ck)(i+j-3k)=0$

$a+b-3c=0\to a+b=3c\to a=3c-b$

$(ai+bj+ck)(i+3j-2k)=5$

$+3b-2c=5$

$3c-b+3b-2c=5$

$2b+c=5\to c=5-2b$

$(ai+bj+ck)(2i+j+4k)=8$

$2a+b+c4=8$

$2a+b+4c=8$

$2(3c-b)+b+4(5-2b)=8$

$6(5-2b)-2b+b+20-8b=8$

$30-12b-b+20-8b=8$

$30-12b-b+20-8b=8$

$42=21b$

$b=2$

$c=5-2\left(2\right)=5-4=1$

$a=3c-b=3\left(1\right)-2=1$

$\therefore$ vector is $i+2j+k$