1.$\bar{a}=-9 \hat{i}+6 \hat{j}+15 \hat{k} \bar{b}=6 \hat{i}-4 \hat{j}-10 \hat{k}$
2.$\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=5 \hat{i}-2 \hat{j}+4 \hat{k}$
3.$\bar{a}=-\frac{3}{5} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{3} \hat{k}, \bar{b}=5 \hat{i}+4 \hat{j}+3 \hat{k}$
4.$\bar{a}=4 \hat{i}-\hat{j}+6 \hat{k}, \bar{a}=5 \hat{i}-2 \hat{j}+4 \hat{k}$
$=-\frac{3}{2}(6 \hat{i}-4 \hat{j}-19 \hat{k})$
$\therefore \bar{a}=-\frac{3}{2} \bar{b}$
i.e. $\bar{a}$ is a non-zero scalar multiple of $\bar{b}$
Hence, $\bar{a}$ is parallel to $\bar{b}$.
2. $\begin{aligned} & \bar{a} \cdot \bar{b}=(2 \hat{i}+3 \hat{j}-\hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k}) \\ & =(2)(5)+(3)(-2)+(-1)(4)\end{aligned}$ = 10 – 6 – 4 = 0
Since, $\bar{a}, \bar{b}$ are non-zero vectors and $\bar{a} \cdot \bar{b}=0, \bar{a}$ is orthogonal to $\bar{b}$.
3. $\begin{aligned} & \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\left(-\frac{3}{5} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\right) \cdot(5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & =\left(-\frac{3}{5}\right)(5)+\left(\frac{1}{2}\right)(4)+\left(\frac{1}{3}\right)(3)\end{aligned}$
= -3 + 2 + 1 = 0
Since $\bar{a} \cdot \bar{b}$ are non-zero vectors and $\bar{a} \cdot \bar{b}=0$
$\bar{a}$ is orthogonal to $\bar{b}$
4. $\begin{aligned} & \bar{a} \cdot \bar{b}=(4 \hat{i}-\hat{j}+6 \hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k}) \\ & =(4)(5)+(-1)(-2)+(6)(4) \\ & =20+2+24 \\ & =46 \neq 0\end{aligned}$
$\therefore \bar{a}$ is not orthogonal to $\bar{b}$.
It is clear that $\bar{a}$ is not a scalar multiple of $\bar{b}$.
$\therefore \bar{a}$ is not parallel to $\bar{b}$.
Hence, $\bar{a}$ is neither parallel nor orthogonal to $\bar{b}$.
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$\sec \left(\frac{x^5+y^5}{x^5-y^5}\right)=a^2$