If vectors $\vec{a}$ and $\vec{b}$ represent the sides $\vec{AB}$ and $\vec{BC}$ of a regular hexagon $ABCDEF$,find the vector represented by $\vec{FA}$.

  • A
    $\vec{a} + \vec{b}$
  • B
    $\vec{b} - \vec{a}$
  • C
    $\vec{a} - \vec{b}$
  • D
    $2\vec{b} - \vec{a}$

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Similar Questions

Column-$I$Column-$II$
$(A)$ In $R^2$,if the magnitude of the projection vector of the vector $\alpha \hat{i}+\beta \hat{j}$ on $\sqrt{3} \hat{i}+\hat{j}$ is $\sqrt{3}$ and if $\alpha=2+\sqrt{3} \beta$,then possible value$(s)$ of $|\alpha|$ is (are)$(P)$ $1$
$(B)$ Let $a$ and $b$ be real numbers such that the function $f(x)=\begin{cases} -3ax^2-2, & x < 1 \\ bx+a^2, & x \geq 1 \end{cases}$ is differentiable for all $x \in R$. Then possible value$(s)$ of $a$ is (are)$(Q)$ $2$
$(C)$ Let $\omega \neq 1$ be a complex cube root of unity. If $(3-3\omega+2\omega^2)^{4n+3} + (2+3\omega-3\omega^2)^{4n+3} + (-3+2\omega+3\omega^2)^{4n+3}=0$,then possible value$(s)$ of $n$ is (are)$(R)$ $3$
$(D)$ Let the harmonic mean of two positive real numbers $a$ and $b$ be $4$. If $q$ is a positive real number such that $a, 5, q, b$ is an arithmetic progression,then the value$(s)$ of $|q-a|$ is (are)$(S)$ $4$
$(T)$ $5$

Let $A$ and $B$ be two points. The position vector of $A$ is $6b - 2a$. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,what is the position vector of $B$?

If $\vec{a} = \lambda x \hat{i} + y \hat{j} + 4z \hat{k}$,$\vec{b} = y \hat{i} + x \hat{j} + 3y \hat{k}$,and $\vec{c} = -z \hat{i} - 2z \hat{j} - (\lambda + 1) \hat{k}$ are the sides of the triangle $ABC$,where $x, y, z$ are not all zero,such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,then the value of $\lambda$ is:

The minimum value of $(x_1 - x_2)^2 + (\sqrt{2 - x_1^2} - \frac{9}{x_2})^2$ where $x_1 \in (0, \sqrt{2})$ and $x_2 \in R^+$.

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$, then $k$ is equal to

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