Shortcut Methods
JEE Main
- Modulus of a vector:
$$| \vec{a}| = \sqrt{a_{x}^{2} + a_{y}^{2}+ a_{z}^{2}}$$
- Unit vectors:
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$
$$\hat{i} = \frac{\vec{i}}{|\vec{i}|} = (1,0,0)$$
$$\hat{j} = \frac{\vec{j}}{|\vec{j}|} = (0,1,0)$$
$$\hat{k} = \frac{\vec{k}}{|\vec{k}|} = (0,0,1)$$
- Direction Cosines:
The direction cosines of a vector (\vec{a}) with respect to the positive x-axis, y-axis, and z-axis respectively are: $$\cos\alpha = \frac{a_{x}}{|\vec{a}|}, \qquad \cos\beta = \frac{a_{y}}{|\vec{a}|}, \qquad \cos\gamma = \frac{a_{z}}{|\vec{a}|}$$
where (a_{x}), (a_{y}), and (a_{z}) are the components of the vector in the x, y, and z directions, respectively.
- Addition of vectors:
$$\vec{a}+\vec{b}=(a_{x}+b_{x})\hat{i}+(a_{y}+b_{y})\hat{j}+(a_{z}+b_{z})\hat{k}$$
- Subtraction of vectors:
$$\vec{a}-\vec{b}=(a_{x}-b_{x})\hat{i}+(a_{y}-b_{y})\hat{j}+(a_{z}-b_{z})\hat{k}$$
- Scalar product of vectors:
$$\vec{a}\cdot\vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}$$
- Vector product of vectors:
$$\vec{a}\times\vec{b}=\left|\begin{array} \hat{i} & \hat{j} & \hat{k} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{array}\right| = (a_{y}b_{z} - a_{z}b_{y})\hat{i} +(a_{z}b_{x} - a_{x}b_{z})\hat{j} +(a_{x}b_{y} - a_{y}b_{x})\hat{k}$$
- Angle between two vectors:
$$\theta =\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{ |\vec{a}||\vec{b}|}\right)$$
CBSE Board Exam
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Modulus of a vector: $$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$$
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Unit vector: $$\hat{a}=\frac{\vec{a}}{|\vec{a}|}$$
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Direction cosines: $$\cos \alpha=\frac{a_1}{|\vec{a}|, }\cos \beta=\frac{a_2}{|\vec{a}|},\cos \gamma=\frac{a_3}{|\vec{a}|}$$
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Addition of vectors: $$\vec{a}+\vec{b}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k}$$
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Subtraction of vectors: $$\vec{a}-\vec{b}=(a_1-b_1)\hat{i}+(a_2-b_2)\hat{j}+(a_3-b_3)\hat{k}$$
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Scalar product of vectors: $$\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3$$
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Angle between two vectors: $$\theta=\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\ |\vec{b}|} \right)\alpha$$