If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$, then ${(a+3b) \times (3a-b)}^2$ is equal to

  • A
    $425$
  • B
    $375$
  • C
    $325$
  • D
    $300$

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Similar Questions

If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is:

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?

If $a = (1, 1, 1)$ and $c = (0, 1, -1)$ are two vectors and $b$ is a vector such that $a \times b = c$ and $a \cdot b = 3$,then $b$ is equal to

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{(\bar{b}+\bar{c})}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Let the position vectors of two points $P$ and $Q$ be $3 \hat{i} - \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} - 4 \hat{k}$ respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4, -1, 2)$ and $(-2, 1, -2)$ respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\vec{TA}$ is perpendicular to both $\vec{PR}$ and $\vec{QS}$ and the length of vector $\vec{TA}$ is $\sqrt{5}$ units,then the modulus of a position vector of $A$ is

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