Question

The vector $a,b$ and $c$ are equal in length and taken pairwise, they make equal angles. If $a=i+j,b=j+k,$ and $c$ makes an obtuse angle with the base vector $i$ and $c$ is equal to

• A. $i+k$
• B. $-i+4j-k$
• C. $-\frac{1}{3}i+\frac{4}{3}j-\frac{1}{3}k$
• D. $\frac{1}{3}i-\frac{4}{3}j+\frac{1}{3}k$

Answer 1

Answer: (C)

$-\frac{1}{3}i+\frac{4}{3}j-\frac{1}{3}k$

The length of $a$ is $\left|a\right|=\sqrt{1+1}=\sqrt{2}$

Similarly $\left|b\right|=\sqrt{1+1}=\sqrt{2}$

Since the three vectors have equal lengths $\left|c\right|=\sqrt{2}$

Let $c=c_{1}i+c_{2}j+c_{3}k$

Since $c$ makes an abtuse angle with $i,$ then $c.i=c_1<0$

We are also given that the angle between the vectors are qual so

${\cos }^{-1}\frac{a.b}{\left|a\right|\left|b\right|}={\cos }^{-1}\frac{a.c}{\left|a\right|\left|c\right|}={\cos }^{-1}\frac{b.c}{\left|c\right|\left|b\right|}$

Now $a.b=1,a.c=c_1+c_2,b.c=c_3+c_2$

${\cos }^{-1}\frac{a.b}{\left|a\right|\left|b\right|}={\cos }^{-1}\frac{1}{2}={\cos }^{-1}\frac{c_1+c_2}{2}={\cos }^{-1}\frac{c_3+c_2}{2}$

This gives $c_1+c_2=1$ and $c_3+c_2=1$ this gives

$c_2=1-c_1$ and $c_3=c_1$

Substitute these values we get $2={c_1}^2+{c_2}^2+{c_3}^2\Rightarrow 2={c_1}^2+{(1-c_1)}^2+{c_1}^2$

This on solving gives $c_1=1,\frac{-1}{3}$

Since $c_1$ must be less than zero, we rule out the positive value obtained.

Thus $c_1=c_3=\frac{-1}{3}$ and $c_2=1-c_1=\frac{4}{3}$

Hence the required vector is $c=\frac{-1}{3}i+\frac{4}{3}j-\frac{-1}{3}k$