91影视

To find the direction cosines we will use the formula \(cos\theta=\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\).

Let the vector \(u=\left<1,2,3 \right>\).

So, the angle between the x-axis and the vector \(u\) is \(cos\alpha =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<1,0,0 \right>\).

Therefore,

\(cos\alpha =\frac{\left<1,2,3 \right>\cdot \left<1,0,0 \right>}{\sqrt{1^{2}+2^{2}+3^{2}}\sqrt{1^{2}+0+0}}\)

\(cos\alpha =\frac{1(1)+2(0)+3(0)}{\sqrt{1+4+9}\sqrt{1}}\)

\(cos\alpha =\frac{1+0+0}{\sqrt{14}\sqrt{1}}\)

\(cos\alpha =\frac{1}{\sqrt{14}}\)

Now, the angle between the y-axis and the vector \(u\) is \(cos\beta =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<0,1,0 \right>\).

Therefore,

\(cos\beta=\frac{\left<1,2,3 \right>\cdot \left<0,1,0 \right>}{\sqrt{1^{2}+2^{2}+3^{2}}\sqrt{0+1^{2}+0}}\)

\(cos\beta=\frac{1(0)+2(1)+3(0)}{\sqrt{1+4+9}\sqrt{1}}\)

\(cos\beta=\frac{0+2+0}{\sqrt{14}\sqrt{1}}\)

\(cos\beta=\frac{2}{\sqrt{14}}\)

Now, the angle between the z-axis and the vector \(u\) is \(cos\gamma =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<0,0,1 \right>\).

Therefore,

\(cos\gamma =\frac{\left<1,2,3 \right>\cdot \left<0,0,1 \right>}{\sqrt{1^{2}+2^{2}+3^{2}}\sqrt{0+0+1^{2}}}\)

\(cos\gamma =\frac{1(0)+2(0)+3(1)}{\sqrt{1+4+9}\sqrt{1}}\)

\(cos\gamma =\frac{0+0+3}{\sqrt{14}\sqrt{1}}\)

\(cos\gamma =\frac{3}{\sqrt{14}}\)

Now, the direction angles are,

\(cos\alpha =\frac{1}{\sqrt{14}}, \alpha =cos^{-1}\left ( \frac{1}{\sqrt{14}} \right )\)

\(cos\beta =\frac{2}{\sqrt{14}}, \beta=cos^{-1}\left ( \frac{2}{\sqrt{14}} \right )\)

\(cos\gamma =\frac{3}{\sqrt{14}}, \gamma=cos^{-1}\left ( \frac{3}{\sqrt{14}} \right )\)