A unit vector in the xy-plane which is perpen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the required unit vector in the $xy$-plane be $\vec{r} = xi + yj$.Since $\vec{r}$ is a unit vector, we have $x^2 + y^2 = 1$.Given that $\vec{r

Vedclass · Accepted Answer

$\frac{1}{5}(3i + 4j)$

Vedclass · Answer

(B) Let the required unit vector in the $xy$-plane be $\vec{r} = xi + yj$.Since $\vec{r}$ is a unit vector, we have $x^2 + y^2 = 1$.Given that $\vec{r}$ is perpendicular to $\vec{a} = 4i - 3j + k$, their dot product must be zero:$(xi + yj) \cdot (4i - 3j + k) = 0$$4x - 3y = 0 \Rightarrow 4x = 3y \Rightarrow x = \frac{3}{4}y$.Substituting $x = \frac{3}{4}y$ into $x^2 + y^2 = 1$:$(\frac{3}{4}y)^2 + y^2 = 1$$\frac{9}{16}y^2 + y^2 = 1 \Rightarrow \frac{25}{16}y^2 = 1$$y^2 = \frac{16}{25} \Rightarrow y = \pm \frac{4}{5}$.If $y = \frac{4}{5}$, then $x = \frac{3}{4} 	imes \frac{4}{5} = \frac{3}{5}$.Thus, the vector is $\frac{3}{5}i + \frac{4}{5}j = \frac{1}{5}(3i + 4j)$.Comparing this with the given options, the correct option is $B$.

$A$ unit vector in the $xy$-plane which is perpendicular to $4i - 3j + k$ is

Similar Questions

If $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,$|\vec{a}|=3$,$|\vec{b}|=5$,and $|\vec{c}|=7$,then the angle between $\vec{a}$ and $\vec{b}$ is

If two vectors $\vec{a}$ and $\vec{b}$ which are perpendicular to each other are such that $|\vec{a}|=8$ and $|\vec{b}|=3$,then $|\vec{a}-2\vec{b}|=$

Let $a = 2i - j + k$,$b = i + 2j - k$,and $c = i + j - 2k$ be three vectors. $A$ vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{2/3}$ is

If the resultant of two forces is of magnitude $P$ and equal to one of them and perpendicular to it,then the other force is

Vedclass Test Series

Exam Paper Generator

Online Exam Module