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Geometrical Interpretation of Product of Vectors - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Geometrical Interpretation of Vector product is considered one of the most asked concept.

  • 12 Questions around this concept.

Solve by difficulty

QuestionMEDIUM

A particle is acted upon by constant forces 4\hat{i}+\hat{j}-3\hat{k}\; and\; 3\hat{i}+\hat{j}-\hat{k}\; which displace it from a point \hat{i}+2\hat{j}+3\hat{k}\; to the point 5\hat{i}+4\hat{j}+\hat{k}\; . The work done in standard units by the forces is given by

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Let $\vec{a}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$, Let $\vec{\beta}_1$ be parallel to $\vec{a}$ and $\vec{\beta}_2$ be perpendicular to $\vec{a}$.If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})_{\text {is }}$

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If \overrightarrow{\mathrm{a}}=\hat{\imath}+2 \hat{k}, \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}, \vec{c}=7 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}, \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{0} and  \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0.
Then \vec{r} \cdot \vec{c} is equal to

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Let for a triangle ABC,

\begin{aligned} & \overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{CB}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\delta \hat{\mathrm{k}} \end{aligned}

If  \delta>0 and the area of the triangle  \mathrm{ABC}  is  5 \sqrt{6} , then \overrightarrow{\mathrm{CB}} \cdot \overrightarrow{\mathrm{CA}}  is equal to

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Concepts Covered - 1

Geometrical Interpretation of Vector product

Area of Parallelogram

If \vec {\mathbf a} and \vec {\mathbf b}, are two non-zero, non-parallel vectors represented by AD and AB respectively and let Ө be the angle between them.

\\\mathrm{In\;\;\Delta\;ADE,\;\;}\sin\theta=\frac{DE}{AD}\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;DE=AD \sin \theta=|\vec{\mathbf{a}}| \sin \theta\\\text { Area of parallelogram } \mathrm{ABCD}=\mathrm{AB} \cdot \mathrm{DE}\\\text{Thus,}\\\text { Area of parallelogram } \mathrm{ABCD}=|\vec{b} \| \vec{a}| \sin \theta=|\vec{a} \times \vec{b}|

 

Area of Triangle

If \vec {\mathbf a} and \vec {\mathbf b}, are two non-zero, non-parallel vectors represent as the adjacent sides of a triangle then its area is given as \frac{1}{2}\left | \vec {\mathbf a}\times\vec {\mathbf b} \right |

The area of a triangle is ½ (Base) x (Height)

From the figure,

\\\text { Area of triangle } \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \cdot \mathrm{CD}\\\\\text { But } \mathrm{AB}=|\vec{\mathbf b}| \text { (as given), and } \mathrm{CD}=|\vec{\mathbf a}| \sin \theta\\\\\text { Thus, Area of triangle } \mathrm{ABC}=\frac{1}{2}|\vec{\mathbf b}||\vec{\mathbf a}| \sin \theta=\frac{1}{2}|\vec{\mathbf a} \times \vec{\mathbf b}|

 

NOTE:

\\1.\;\;\;\text {The area of a parallelogram with diagonals } \vec{\mathbf{d}}_{1} \text { and } \vec{\mathbf{d}}_{2}\;\;\text{is }\;\frac{1}{2}\left |\vec{\mathbf{d}}_{1}\times \vec{\mathbf{d}}_{2} \right |.\\\\2.\;\;\;\text {The area of a plane quadrilateral } A B C D \text { with AC and BD as diagonal }\\\mathrm{\;\;\;\;\;}\text{ is}\;\;\frac{1}{2}\left | \overrightarrow{\mathbf {AC}}\times \overrightarrow{\mathbf {BD}} \right |.\\\\3.\;\;\;\text{If }\;\vec{\mathbf a},\;\vec{\mathbf b}\;\text{and }\;\vec{\mathbf c}\text { are position vectors of a } \Delta A B C, \text { then its area is}\\\mathrm{\;\;}\mathrm{\;\;\;}\frac{1}{2}\left | (\vec{\mathbf a}\times \vec{\mathbf b})+(\vec{\mathbf b}\times \vec{\mathbf c})+(\vec{\mathbf c}\times \vec{\mathbf a}) \right |.

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Geometrical Interpretation of Vector product

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Geometrical Interpretation of Vector product

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 3.22

Line : 12

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