MATH 127
Vector Analysis
Triple Product
1. \(\vec{A} .(\vec{B} \times \vec{C} )\)
2. \(\vec{A} \times (\vec{B} \times \vec{C} )\)
3. \((\vec{A} .\vec{B} ). \vec{C} \)
• Projection:
i) Scaler Projection: $\displaystyle \vec{b} \frac{\vec{a} }{|\vec{a}| } $
ii) Vector Projection: $\displaystyle \frac{\vec{a} .\vec{b} }{|\vec{a}|^2} \vec{a} $
The scaler projection of
\(\vec{b} \)
and
\(\vec{a} \)
is the length of the AB (shown in the figure). The vector projection of
\(\vec{b} \)
onto
\(\vec{a} \)
is the vector with this length at the point A point in the same direction as
\(\vec{a} \)
.
This quantity is also called the component of
\(\vec{b} \)
in the
\(\vec{a} \)
direction. And, the vector projection is merely the unit vector
$ \displaystyle \frac{\vec{a} }{|\vec{a} } $
times the scalar projection of
\(\vec{b} \)
onto
\(\vec{a} \)
.
\(\therefore\) The scalar projection of \(\vec{b} \) onto \(\vec{a} \) is proj $ \displaystyle \frac{\vec{b} }{\vec{a}} = \vec{b} .\frac{\vec{a}}{\vec{a}} $
and the vector projection of \(\vec{b} \) onto \(\vec{a} \) is proj $ \displaystyle \frac{\vec{b} }{\vec{a}} = \frac{\vec{a} .\vec{b} }{|\vec{a} |^2} \vec{a} $ .
♦ Direction Cosines:
$$\cos \alpha = \frac{x}{|\vec{r}|} = \frac{x}{\sqrt{ x^2 + y^2 + z^2}}$$
$$\cos \beta = \frac{y}{|\vec{r}|} = \frac{y}{\sqrt{x^2 + y^2 + z^2}} $$
$$\cos \gamma = \frac{z}{|\vec{r}|} = \frac{z }{\sqrt{x^2 + y^2 + z^2} }$$
Theorem: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
Example: Find direction cosines of a vector represented by
\( \vec{PQ }\)
where
\(P(2,- 3, 5) \textrm{ and } Q(1,0,- 1 )\)
are two point.
Solution: \begin{align*} \vec{PQ} & = (1 - 2) \hat{i} + (0 + 3) \hat{j} + (- 1 - 5) \hat{k} \\ & = - \hat{i} + 3 \hat{j} - 5 \hat{k} \\ \textrm{Now,} & \\ \alpha & = \cos^{-1} \left( \frac{- 1}{\sqrt[]{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \left(- \frac{1}{\sqrt{46}}\right) ~= 98.47^\circ \\ \beta & = \cos^{-1} \left(\frac{3}{\sqrt{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \frac{3}{\sqrt{46}} ~= 63.74^\circ \\ \theta & = \cos^{-1} \left(\frac{- 6}{\sqrt[]{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \frac{6}{\sqrt{46}} ~= 152.2^\circ \end{align*}
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