Question

a, b, c are unit vectors such that a and b are mutually perpendicualr ans c is equally inclined to a and b at an angle$\theta .$ If c$=xa+yb+z(a\times b),$ then:

• A. $z^2=1-2y^2$
• B. $z^2=1-x^2-y^2$
• C. $z^2=1-2x^2$
• D. $x^2=y^2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $z^2=1-2y^2$
B $z^2=1-x^2-y^2$
C $z^2=1-2x^2$
D $x^2=y^2$

Given $a.b=0$ and $c.a=c.b=1.1\cos \theta$
$\therefore a\times b=\mathrm{1.1.}\sin {90}^{\circ }n$ $\therefore {(a\times b)}^3=1$
Also we know that $a.(a\times b)=\left[aab\right]=0,b.(a\times b)=0$ and $c.(a\times b)=\left[cab\right]=\left[abc\right]$
Now, $c=xa+yb+z(a\times b)$
Multiplying both sides scalarly by a and b $\cos \theta =x.1+0+0,\cos \theta =y.1$ ...(1)
Again multiplying by $a\times b$ $c.(a\times b)=x.0+y.0+z{(a\times b)}^2=z.1$ $\left[abc\right]=z$ Now ${\left[abc\right]}^2=\left|\begin{array}{ccc}a.a& a.b& a.c\\ b.a& b.b& b.c\\ c.a& c.b& c.c\end{array}\right|$ or $z^2=\left|\begin{array}{ccc}1& 0& \cos \theta \\ 0& 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right|=1-2{\cos }^2\theta$ $\therefore z^2=1-2x^2=1-2y^2$ by (1) $=1-x^2-x^2=1-x^2-y^2.$
Also $x^2=y^2.$