a) Show that the vector a b is perpendicular to vectors a and b . [2] b) Evaluate: dx1 2 x . [2] c) Evaluate, using L' Hopital rule: x 0 e^x x 1x^2 . [2]

a) Show that the vector $\vec{a}\times \vec{b}$ is perpendicular to vectors $\vec{a}$ and $\vec{b}$.

Proof:

Let $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$ be two vectors in ${\mathbb{R}}^3$. The cross product $\vec{a}\times \vec{b}$ is defined as:

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\mathbf{i}-(a_1{b}_3-a_3{b}_1)\mathbf{j}+(a_1{b}_2-a_2{b}_1)\mathbf{k}$$

Let $\vec{c}=\vec{a}\times \vec{b}=(c_1,c_2,c_3)$, where: $$c_1=a_2{b}_3-a_3{b}_2,c_2=-(a_1{b}_3-a_3{b}_1),c_3=a_1{b}_2-a_2{b}_1.$$

Perpendicularity to $\vec{a}$:

To show that $\vec{c}$ is perpendicular to $\vec{a}$, we compute the dot product $\vec{a}\cdot \vec{c}$:

$$\vec{a}\cdot \vec{c}=a_1{c}_1+a_2{c}_2+a_3{c}_3$$

Substitute the values of $c_1,c_2,c_3$:

$$\vec{a}\cdot \vec{c}=a_1(a_2{b}_3-a_3{b}_2)+a_2[-(a_1{b}_3-a_3{b}_1)]+a_3(a_1{b}_2-a_2{b}_1)$$

Simplify:

$$=a_1{a}_2{b}_3-a_1{a}_3{b}_2-a_1{a}_2{b}_3+a_2{a}_3{b}_1+a_1{a}_3{b}_2-a_2{a}_3{b}_1$$

$$=(a_1{a}_2{b}_3-a_1{a}_2{b}_3)+(-a_1{a}_3{b}_2+a_1{a}_3{b}_2)+(a_2{a}_3{b}_1-a_2{a}_3{b}_1)=0$$

Since $\vec{a}\cdot \vec{c}=0$, $\vec{c}$ is perpendicular to $\vec{a}$.

Perpendicularity to $\vec{b}$:

Similarly, compute $\vec{b}\cdot \vec{c}$:

$$\vec{b}\cdot \vec{c}=b_1{c}_1+b_2{c}_2+b_3{c}_3$$

Substitute the values of $c_1,c_2,c_3$:

$$\vec{b}\cdot \vec{c}=b_1(a_2{b}_3-a_3{b}_2)+b_2[-(a_1{b}_3-a_3{b}_1)]+b_3(a_1{b}_2-a_2{b}_1)$$

Simplify:

$$=a_2{b}_1{b}_3-a_3{b}_1{b}_2-a_1{b}_2{b}_3+a_3{b}_1{b}_2+a_1{b}_2{b}_3-a_2{b}_1{b}_3$$

$$=(a_2{b}_1{b}_3-a_2{b}_1{b}_3)+(-a_3{b}_1{b}_2+a_3{b}_1{b}_2)+(-a_1{b}_2{b}_3+a_1{b}_2{b}_3)=0$$

Since $\vec{b}\cdot \vec{c}=0$, $\vec{c}$ is perpendicular to $\vec{b}$.

Thus, $\vec{a}\times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

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b) Evaluate: $\int \frac{dx}{1-2\cos x}$.

Solution:

We use the substitution method. Let:

$$I=\int \frac{dx}{1-2\cos x}$$

Recall the trigonometric identity:

$$1-2\cos x=-2{\cos }^2\frac{x}{2}+2{\sin }^2\frac{x}{2}+1-1=2{\sin }^2\frac{x}{2}-2{\cos }^2\frac{x}{2}+1-1$$

Actually, a better approach is to use the substitution $t=\tan \frac{x}{2}$ (Weierstrass substitution):

Let $t=\tan \frac{x}{2}$, then:

$$\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2}{1+t^2}dt$$

Substitute into the integral:

$$I=\int \frac{1}{1-2\left(\frac{1-t^2}{1+t^2}\right)}\cdot \frac{2}{1+t^2}dt$$

Simplify the denominator:

$$1-2\left(\frac{1-t^2}{1+t^2}\right)=\frac{(1+t^2)-2(1-t^2)}{1+t^2}=\frac{1+t^2-2+2t^2}{1+t^2}=\frac{3t^2-1}{1+t^2}$$

Thus, the integral becomes:

$$I=\int \frac{1+t^2}{3t^2-1}\cdot \frac{2}{1+t^2}dt=2\int \frac{1}{3t^2-1}dt$$

Factor the denominator:

$$3t^2-1=(\sqrt{3}t{)}^2-1=(\sqrt{3}t-1)(\sqrt{3}t+1)$$

Rewrite the integral:

$$I=2\int \frac{1}{(\sqrt{3}t-1)(\sqrt{3}t+1)}dt$$

Use partial fractions:

$$\frac{1}{(\sqrt{3}t-1)(\sqrt{3}t+1)}=\frac{A}{\sqrt{3}t-1}+\frac{B}{\sqrt{3}t+1}$$

Solve for $A$ and $B$:

$$1=A(\sqrt{3}t+1)+B(\sqrt{3}t-1)$$

Let $t=\frac{1}{\sqrt{3}}$:

$$1=A(1+1)+B(0)⟹2A=1⟹A=\frac{1}{2}$$

Let $t=-\frac{1}{\sqrt{3}}$:

$$1=A(0)+B(-1-1)⟹-2B=1⟹B=-\frac{1}{2}$$

Thus:

$$I=2\int \left(\frac{1/2}{\sqrt{3}t-1}-\frac{1/2}{\sqrt{3}t+1}\right)dt=\int \left(\frac{1}{\sqrt{3}t-1}-\frac{1}{\sqrt{3}t+1}\right)dt$$

Integrate:

$$I=\frac{1}{\sqrt{3}}\ln |\sqrt{3}t-1|-\frac{1}{\sqrt{3}}\ln |\sqrt{3}t+1|+C$$

Combine the logarithms:

$$I=\frac{1}{\sqrt{3}}\ln \left|\frac{\sqrt{3}t-1}{\sqrt{3}t+1}\right|+C$$

Substitute back $t=\tan \frac{x}{2}$:

$$I=\frac{1}{\sqrt{3}}\ln \left|\frac{\sqrt{3}\tan \frac{x}{2}-1}{\sqrt{3}\tan \frac{x}{2}+1}\right|+C$$

Final Answer:

$$\boxed{\frac{1}{\sqrt{3}}\ln \left|\frac{\sqrt{3}\tan \frac{x}{2}-1}{\sqrt{3}\tan \frac{x}{2}+1}\right|+C}$$

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c) Evaluate, using L'Hôpital's rule: ${\lim }_{x\to 0}\frac{e^x-x-1}{x^2}$.

Solution:

The given limit is of the indeterminate form $\frac{0}{0}$ when $x\to 0$:

$$\underset{x\to 0}{\lim }\frac{e^x-x-1}{x^2}=\frac{e^0-0-1}{0}=\frac{0}{0}$$

Apply L'Hôpital's rule by differentiating the numerator and denominator:

Differentiate the numerator $N(x)=e^x-x-1$:

$$N^{\prime }(x)=e^x-1$$

Differentiate the denominator $D(x)=x^2$:

$$D^{\prime }(x)=2x$$

Now, the limit becomes:

$$\underset{x\to 0}{\lim }\frac{e^x-1}{2x}$$

Again, this is of the form $\frac{0}{0}$ when $x\to 0$, so apply L'Hôpital's rule again:

Differentiate the numerator $N^{\prime }(x)=e^x-1$:

$$N^{\prime \prime }(x)=e^x$$

Differentiate the denominator $D^{\prime }(x)=2x$:

$$D^{\prime \prime }(x)=2$$

Now, the limit is:

$$\underset{x\to 0}{\lim }\frac{e^x}{2}=\frac{e^0}{2}=\frac{1}{2}$$

Final Answer:

$$\boxed{\frac{1}{2}}$$