let the vector Q is given as
[tex]\vec Q = a\hat i + b\hat j + c\hat k[/tex]
given that
[tex]P X Q = -6\hat j[/tex]
here we know that
[tex]P = 4\hat i + 3 \hatk[/tex]
now by above equation
[tex](4\hat i + 3\hat k) X (a\hat i + b\hat j + c\hat k) = - 6\hat j[/tex]
[tex]4b\hat k - 4c\hat j + 3a\hat j - 3b\hat i = - 6\hat j[/tex]
so by comparing both sides
b = 0
4c - 3a = 6
also we know that
[tex]a^2 + b^2 + c^2 = 17^2[/tex]
[tex]a^2 + 0 + (1.5 + 0.75a)^2 = 289[/tex]
by solving above equation
a = 12.85 and c = 11.14
so the vector Q is given as
[tex]Q = 12.85\hat i + 11.14\hat k[/tex]
