Overrightarrow A = 2hat{i} + 4hat{j} + 4hat{k} and overrightarrow B = 4hat{i} + 2hat{j}

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3-1.Vectors

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$\overrightarrow A = 2\hat i + 4\hat j + 4\hat k$ and $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ are two vectors. The angle between them will be ........ $^o$

$0$

$45$

$90$

$60$

Solution

(c) $\cos \theta = \frac{{\overrightarrow {\rm A} \,.\,\overrightarrow B }}{{|\overrightarrow {\rm A} \,|\,.\,|\,\overrightarrow B |}} = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{|\overrightarrow {\rm A} \,|\,.\,|\overrightarrow B |\,}}$

$ = \frac{{2 \times 4 + 4 \times 2 – 4 \times 4}}{{|\overrightarrow A |\,.\,|\overrightarrow B |}} = 0$

$\therefore \theta = {\cos ^{ – 1}}(0^\circ )$ $ \Rightarrow \,\,\,\theta = 90^\circ $

Standard 11

Physics

If diagonals of a parallelogram are $\left( {5\hat i – 4\hat j + 3\hat k} \right)$ and $\left( {3\hat i + 2\hat j – \hat k} \right)$ then its area is

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${\vec A }$, ${\vec B }$ and ${\vec C }$ are three non-collinear, non co-planar vectors. What can you say about directin of $\vec A \, \times \,\left( {\vec B \, \times \vec {\,C} } \right)$ ?

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If $F _1$ and $F _2$ are two vectors of equal magnitudes $F$ such that $\left| F _1 \cdot F _2\right|=\left| F _1 \times F _2\right|$, then $\left| F _1+ F _2\right|$ equals to

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The two vectors $\vec A$ and $\vec B$ that are parallel to each other are

medium

$\overrightarrow A = 2\hat i + 4\hat j + 4\hat k$ and $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ are two vectors. The angle between them will be ........ $^o$

Solution

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