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Vector (or Cross) Product of Two Vectors is considered one the most difficult concept.
Vector Product in Terms of Components is considered one of the most asked concept.
79 Questions around this concept.
Let and . If is a unit vector such that and then is equal to:
If and the angle between is equal to
are two vectors and is a vector such that then
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Let the position vector S of point A and B be $\hat{i}+\widehat{j}+\widehat{k}$ and $\widehat{2 i}-\widehat{j}-\widehat{3 k}$ respectively. A point 'p' divides the line segment AB intervally in the ratio $\lambda: 1(\lambda, 0)$ if $O$ is the origin and $\overrightarrow{O B} \cdot \overrightarrow{O P}-3|\overrightarrow{O A} \times \overrightarrow{O P}|^2=6$ then $\lambda$ is equal to $\qquad$
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to
Let $\overrightarrow{\mathrm{a}}=-5 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{\mathrm{i}}) \times \hat{\mathrm{i}}) \times \hat{\mathrm{i}}$. Then $\overrightarrow{\mathrm{c}} \cdot(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ is equal to :
Vector product is :
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$(3\hat{i}\times 9\hat{j})+(7\hat{k}\times 3\hat{k})=$
$\hat{i}\times (2\hat{i}\times \hat{j})$ is along which axis :
$5 \hat{i} \times 4 \hat{j}=?$
The vector product of two nonzero vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, is denoted by $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ and defined as,
$
\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \sin \theta \hat{\mathbf{n}}
$
where $\theta$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}, 0 \leq \theta \leq \pi$ and $\hat{\mathbf{n}}$ is a unit vector perpendicular to both $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, such that $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\hat{\mathbf{n}}$ form a right hand system.
Observe that, the direction of $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ is opposite to that of $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}$ as shown in the figure.
i.e. $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$
Properties of Vector Product
1. $\vec{a} \times \vec{b}$ is a vector.
2. Let $\vec{a}$ and $\vec{b}$ be two nonzero vectors. Then $\vec{a} \times \vec{b}=\overrightarrow{0}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel (or collinear) to each other, i.e.,
$
\vec{a} \times \vec{b}=\overrightarrow{0} \Leftrightarrow \vec{a} \| \vec{b}
$
In particular, $\vec{a} \times \vec{a}=\overrightarrow{0}$ and $\vec{a} \times(-\vec{a})=\overrightarrow{0}$, since in the first situation, $\theta=0$ and in the second one, $\theta=\pi$.
3. If $\theta=\frac{\pi}{2}$, then $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}|$.
4. From the property 2 . and 3 .
$
\begin{aligned}
& \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0} \\
& \hat{i} \times \hat{j}=\hat{k}, \quad \hat{j} \times \hat{k}=\hat{i}, \quad \hat{k} \times \hat{i}=\hat{j} \\
& \text { and also, } \\
& \hat{j} \times \hat{i}=-\hat{k}, \quad \hat{k} \times \hat{j}=-\hat{i} \text { and } \quad \hat{i} \times \hat{k}=-\hat{j}
\end{aligned}
$
5. Vector product is not associative,
$
\text { i.e. } \quad \vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}
$
6. If $\vec{a}$ and $\vec{b}$ are two vectors and $m$ is a scalar, then
$
m \vec{a} \times \vec{b}=m(\vec{a} \times \vec{b})=\vec{a} \times m \vec{b}
$
7. If $\vec{a}$ and $\vec{b}$ are two vectors and $m, n$ are scalars, then $m \vec{a} \times n \vec{b}=m n(\vec{a} \times \vec{b})=m(n \vec{a} \times \vec{b})=n(m \vec{a} \times \vec{b})$
8. For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$, we have
(i) $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$
(ii) $\vec{a} \times(\vec{b}-\vec{c})=\vec{a} \times \vec{b}-\vec{a} \times \vec{c}$
If $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=b_1 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+b_3 \hat{\mathbf{k}}$, then their cross product given by
$
\vec{a} \times \vec{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|
$
Proof:
$
\begin{aligned}
\vec{a} \times \vec{b}= & \left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \\
= & a_1 b_1(\hat{i} \times \hat{i})+a_1 b_2(\hat{i} \times \hat{j})+a_1 b_3(\hat{i} \times \hat{k})+a_2 b_1(\hat{j} \times \hat{i}) \\
& +a_2 b_2(\hat{j} \times \hat{j})+a_2 b_3(\hat{j} \times \hat{k})+a_3 b_1(\hat{k} \times \hat{i})+a_3 b_2(\hat{k} \times \hat{j})+a_3 b_3(\hat{k} \times \hat{k}) \\
= & a_1 b_2(\hat{i} \times \hat{j})-a_1 b_3(\hat{k} \times \hat{i})-a_2 b_1(\hat{i} \times \hat{j}) \\
& ++a_2 b_3(\hat{j} \times \hat{k})+a_3 b_1(\hat{k} \times \hat{i})-a_3 b_2(\hat{j} \times \hat{k})
\end{aligned}
$
Using the fact that,
$
\begin{aligned}
& \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0 \\
\text { and, } & \hat{i} \times \hat{k}=-\hat{k} \times \hat{i}, \hat{j} \times \hat{i}=-\hat{i} \times \hat{j} \text { and } \hat{k} \times \hat{j}=-\hat{j} \times \hat{k} \\
= & a_1 b_2 \hat{k}-a_1 b_3 \hat{j}-a_2 b_1 \hat{k}+a_2 b_3 \hat{i}+a_3 b_1 \hat{j}-a_3 b_2 \hat{i} \\
(\text { as } \hat{i} \times & \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} \text { and } \times k+\hat{j}) \\
= & \left(a_2 b_3-a_3 b_2\right) \hat{i}-\left(a_1 b_3-a_3 b_1\right) \hat{j}+\left(a_1 b_2-a_2 b_1\right) \hat{k} \\
= & \left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|
\end{aligned}
$
The angle between Two Vectors
If $\Theta$ is the angle between the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, then
$
\sin \theta=\frac{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}
$
Vector perpendicular to the plane of two given vectors
The unit vector perpendicular to the plane of $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\frac{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})}{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|}$.
Also note that $-\frac{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})}{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|}$ is also a unit vector perpendicular to the plane of $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$.
Vectors of magnitude ' $\lambda$ ' perpendicular to the plane of $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are given by
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