Question

Let $a=i-j,b=i+k$, find a unit vector c perpendicular to a and coplanar with a and b. also find a unit vector d perpendicular to both a and c.

• A. $c=\frac{1}{\sqrt{6}}(i+j+2k).d=\frac{1}{\sqrt{3}}(i+j-k)$
• B. $c=\frac{1}{\sqrt{6}}(i-j-2k).d=\frac{1}{\sqrt{2}}(i+j+k)$
• C. $c=\frac{1}{\sqrt{6}}(i+j-2k).d=\frac{1}{\sqrt{2}}(i-j-k)$
• D. $c=\frac{1}{\sqrt{6}}(i-j-2k).d=\frac{1}{\sqrt{2}}(i-j-k)$

Answer 1

The correct option is A $c=\frac{1}{\sqrt{6}}(i+j+2k).d=\frac{1}{\sqrt{3}}(i+j-k)$

Let us understand the basic concept to solve this type of problem

Given vectors

$\overrightarrow{a}=\hat{i}-\hat{j}$

$\overrightarrow{b}=\hat{i}+\hat{k}$

Let any unit vector be c

$\overrightarrow{c}=\frac{1}{\sqrt{a^2+b^2+c^2}}(a\hat{i}+b\hat{k}+c\hat{k})$

Given

$\overrightarrow{a}\cdot \overrightarrow{c}=0$

$(\hat{i}-\hat{j})\cdot \left(\frac{1}{\sqrt{a^2+b^2+c^2}}\right(a\hat{i}+b\hat{k}+c\hat{k}\left)\right)=0$

$\frac{a}{\sqrt{a^2+b^2+c^2}}-\frac{b}{\sqrt{a^2+b^2+c^2}}=0$

$a=b$

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\frac{1}{\sqrt{a^2+b^2+c^2}}\left|\begin{array}{ccc}1& -1& 0\\ 1& 0& 1\\ a& b& c\end{array}\right|$=0$

$1\left(b\right)-1(c-a)=0$

$a-c+a=0$ here a=b

$c=2a$

putting b,c in $\overrightarrow{c}$

$\overrightarrow{c}=\frac{1}{\sqrt{a^2+a^2+4a^2}}(a\hat{i}+a\hat{j}+2a\hat{k})$

$\overrightarrow{c}=\frac{1}{\sqrt{6a^2}}(a\hat{i}+a\hat{j}+2a\hat{k})$

HERE option (A) is satisfying all condition of $\overrightarrow{c}$

$\overrightarrow{c}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})$

now let

$\overrightarrow{d}=\frac{1}{\sqrt{x^2+y^2+z^2}}(x\hat{i}+y\hat{j}+z\hat{k})$

$\overrightarrow{a}\cdot \overrightarrow{d}=0$

$(\hat{i}-\hat{j})\cdot \left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right(x\hat{i}+y\hat{j}+z\hat{k}\left)\right)=0$

$\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2+z^2}}=0$

$x=y$

$\overrightarrow{c}\cdot \overrightarrow{d}=0$

$\left(\frac{1}{\sqrt{6}}\right(\hat{i}+\hat{j}+2\hat{k}\left)\right)\cdot \left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right(x\hat{i}+y\hat{j}+z\hat{k}\left)\right)=0$

$\frac{x}{\sqrt{6(x^2+y^2+z^2)}}+\frac{y}{6\left(\sqrt{x^2+y^2+z^2}\right)}+\frac{z}{6\left(\sqrt{x^2+y^2+z^2}\right)}=0$

since $x=y$

$\frac{x}{\sqrt{6(x^2+x^2+z^2)}}+\frac{x}{6\left(\sqrt{x^2+x^2+z^2}\right)}+\frac{z}{6\left(\sqrt{x^2+x^2+z^2}\right)}=0$

$z=-x$

putting y,z in $\overrightarrow{z}$

$\overrightarrow{d}=\frac{1}{\sqrt{x^2+x^2+(-x{)}^2}}(x\hat{i}+x\hat{j}-x\hat{k})$

$\overrightarrow{d}=\frac{1}{\sqrt{3x^2}}(x\hat{i}+x\hat{j}-x\hat{k})$

$x=1$

$\overrightarrow{d}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})$

So option d is wrong in option (A)