Question

If A vector=3î+4j^ , then find the vector perpendicular to vector A in the plane of A.

Answer 1

Let the vector perpendicular to $\vec{\sf{A}}=\sf{3\,\hat i+4\,\^j}$ be,

$\longrightarrow\vec{\sf{B}}=\sf{x\,\hat i+y\,\^j}$

For two perpendicular vectors, their dot product should be zero.

$\longrightarrow\vec{\sf{A}}\cdot\vec{\sf{B}}=\sf{0}$

$\longrightarrow\sf{\left(3\,\hat i+4\,\^j\right)\cdot\left(x\,\hat i+y\,\^j\right)=0}$

$\longrightarrow\sf{3x+4y=0}$

$\longrightarrow\sf{3x=-4y}$

$\longrightarrow\sf{\dfrac{x}{y}=-\dfrac{4}{3}}$

Or,

$\longrightarrow\sf{x:y=4:-3}$

Let,

• $\sf{x=4k}$

• $\sf{y=-3k}$

where $\sf{k}$ is a non - zero real number.

Hence,

$\longrightarrow\underline{\underline{\vec{\sf{B}}=\sf{k\left(4\,\hat i-3\,\^j\right)}}}$

Magnitude of $\vec{\sf{B}}$ is,

$\longrightarrow\left|\vec{\sf{B}}\right|=\sf{\sqrt{4^2+(-3)^2}}$

$\longrightarrow\left|\vec{\sf{B}}\right|=\sf{5}}$

Direction of $\vec{\sf{B}}$ also depends on sign of $\sf{k.}$ That's why unit vector along $\vec{\sf{B}}$ is,

$\longrightarrow\sf{\^B}=\sf{\pm\left(\dfrac{4}{5}\,\hat i-\dfrac{3}{5}\,\^j\right)}$