Question

Find a vector perpendicular to $\mathrm{i}+2 \mathrm{j}$ magnitude of $3 \sqrt{5}$
A.$3j+6j$
B.$6 i-3 j$
C.$4i-2j$
D.$\widehat{\text{i}}-2\text{j}$

Answer 1

When two vectors are perpendicular, their dot-product equals zero. A vector that is perpendicular to both non-parallel vectors is the cross product of two non-parallel vectors. Take the dot product of the vector and Square the magnitude by squaring both sides. After that Equate both the equation and calculate the value of the resultant vector.

Complete step by step solution:
Consider a plane with two vectors a and b, where the angle between a and b equals and the cross product of a and b equals $\text{ }\\!\\!|\\!\\!\text{ a }\\!\\!|\\!\\!\text{ }\\!\\!|\\!\\!\text{ b }\\!\\!|\\!\\!\text{ sin(}\theta \text{)}$ . Because the cross product is completely reversed when the plane is flipped, it is perpendicular to the plane.
In this question-:
Let a vector, $\mathbf{r}=\mathbf{x i}+\mathbf{y j}$ is perpendicular to $\mathbf{A}=(\mathbf{i}+\mathbf{2 j})$
So, the dot product of $r$ and $A$ must be zero. Because $\cos 90^{\circ}=0$
$\text{r}\text{.A=(xi+yj) }\\!\\!\times\\!\\!\text{ (i+2j)=x+2y=}{{\text{0}}^{{}}}$
According to the question, Magnitude of $r$ is $3 \sqrt{5}$ So, $\left.|\mathbf{r}|=\sqrt{(} \mathbf{x}^{2}+\mathbf{y}^{2}\right)=\mathbf{3} \sqrt{\mathbf{5}}$
Taking square both sides, $\mathrm{x}^{2}+\mathrm{y}^{2}=9 \times 5=45$
$x^{2}+y^{2}=45$
From equation (1), $(-2 y)^{2}+y^{2}=45$
$\Rightarrow 4 y^{2}+y^{2}=45$
$\Rightarrow 5 y^{2}=45$
$\Rightarrow y^{2}=9 \Rightarrow y=3$
$\therefore \mathrm{x}=-2 \mathrm{y}=6$
Hence, $\mathbf{r}=(6 \mathbf{i}-3 \mathbf{j})$
Hence,
Option $(\mathrm{B})$ is the correct answer.

Note: The dot product of two vectors that are perpendicular to each other is zero. The commutative law does not apply to the cross product of two vectors. The additive inverse of two vectors is their cross product. The right-hand rule determines the direction of cross product in this case. By multiplying the magnitudes of two vectors with the cosine of the angle between them, the scalar product is obtained.