Question

Given two vectors are $\hat{i}-\hat{j}$ and $\hat{i}+2\hat{j},$ the unit vector is coplanar with the two vectors and perpendicular to the first. Find the vector?

• A. $+\frac{1}{\sqrt{2}}\,(i+k)$
• B. $+\frac{1}{\sqrt{5}}\,(2i+j)$
• C. $+\frac{1}{\sqrt{2}}\,(i+j)$
• D. $+\frac{1}{\sqrt{2}}\,(2i+3j)$

Answer 1

Answer: (C)

$+\frac{1}{\sqrt{2}}\,(i+j)$

Answer 2

Let given two vectors are $a=i-j$ and $b=i+2j$
Again let third unit vector is c.
$\because$ c is coplanar with a, b
$\therefore$ $c=xa+yb$
$=x(i-j)+y(i+2j)$
$\Rightarrow$ $c=(a+y)i+(-x+2y)j$ ..(i)
Also, c is perpendicular to a $\therefore$
$a.c=0$
$\Rightarrow$ $(i-j).\\{(x+y)i+(-x+2y)j\\}=0$
$\Rightarrow$ $(x+y)-(-x+2y)=0$
$\Rightarrow$ $x+y+x-2y=0$
$\Rightarrow$ $2x-y=0\Rightarrow y=2x$
On putting this value of y in E (i), we get
$c=(x+2x)i+(-x+4x)j$
$\Rightarrow$ $c=3xi+3xj$ ..(ii)
But c is a unit vector. So, $|c|=1$
$\Rightarrow$ ${{(3x)}^{2}}+{{(3x)}^{2}}=1\,\Rightarrow 9{{x}^{2}}+9{{x}^{2}}=1$
$\Rightarrow$ ${{x}^{2}}=\frac{1}{18}\,\,\,\Rightarrow \,\,\,x=\frac{1}{3\sqrt{2}}$
On putting this value of x in E (ii), we get $c=3.\frac{1}{3\sqrt{2}}i+3.\frac{1}{3\sqrt{2}}j\,\,\Rightarrow c=\frac{1}{\sqrt{2}}\,(i+j)$

Coplanar vectors are those in a three-dimensional plane that are parallel to one another. In the same plane, the vectors are perpendicular to one another. Any two random vectors on a plane may always be in line with one another. In a three-dimensional space that can be described in vector form, two lines are coplanar. When the scalar product of three vectors is equal to zero, the vectors are referred to as being planar.

The coplanar line is a well-known concept in three-dimensional geometry. In mathematical theory, three lines that are situated on the same plane are referred to as being coplanar and this property is known as the coplanarity of three vectors. Although straight lines have been used as vector equations, a plane is a two-dimensional geometry that stretches into infinity in three-dimensional space.