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The gradient is $\nabla f=(2x,z,y)$, so its value at the point is $(2,-3,1)$. Since the specified direction is already a unit vector, the directional derivative is its dot product with that gradient:

The perpendicular unit vectors represented by $e^{i\theta }$ and $i{e}^{i\theta }$ point radially outwards and transversely in the increasing-angle direction. Reading off their real coefficients in part (b) gives the signed componentsThese are velocity in polar coordinates and acceleration in polar coordinates. In particular $r\dot{\theta }$ is a transverse speed component; $\dot{\theta }$ alone is an angular velocity. The moving basis, rather than any extra force assumption, accounts for the additional acceleration terms. At the origin the polar directions themselves are undefined, although the Cartesian velocity and acceleration can remain well defined.

The curl gives the vorticity:Since $x\mathbf{i}+y\mathbf{j}=\rho \hat{\mathbf{\rho }}$, the requested form isAt the axis the radial unit vector is undefined, but its coefficient vanishes and the Cartesian expression remains smooth. The radial vorticity reflects the change in rotation rate with height.

The common ratio in the two plane equations can be denoted by $u$. Their intersection then has $x=1+4u$, $y=2+3u$, and $z=-1+2u$. The direction vector $(4,3,2)$ has Euclidean norm $\sqrt{29}$, so the requested unit vector form isSubstituting this expression into both equations gives the same ratio $\lambda /\sqrt{29}$, verifying the entire line.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.

At $(1,2)$, the gradient is $(4,2)$. The requested direction must be normalized to the unit vector $(1,1)/\sqrt{2}$. The directional derivative is thereforeUsing $(1,1)$ without normalization would instead measure change per unit of a differently scaled path parameter.

Since $A^2=-I$, the matrix inverse is $A^{-1}=-A$. The orthogonal matrix condition also gives $A^{-1}=A^{\mathsf{T}}$, so $A^{\mathsf{T}}=-A$. Thus $A$ is skew-symmetric, forcing $a=0$ and $c=-b$. Its columns must be unit vectors, giving $b^2=1$. Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.

The direction must first be represented by a unit vector: $\hat{\mathbf{q}}=(1,1,1)/\sqrt{3}$. The scalar projection of the vector area is thereforeIf a vector projection is intended, multiply this scalar by $\hat{\mathbf{q}}$:

Choose Cartesian coordinates with unit vectors satisfying the right-hand rule, $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$. Put $A$, $B$ and $C$ one unit along the positive $x$, $y$ and $z$ axes, respectively. The following original drawing also includes the vectors needed in the next part.

Two directions within the plane areTheir cross product is $(1,1,1)/4$, so a unit vector normal to the plane isIts negative is equally valid. An equation of a plane through three points in vector form isor, parametrically, $\mathbf{r}=\mathbf{e}+s(\mathbf{f}-\mathbf{e})+t(\mathbf{g}-\mathbf{e})$. Since the coordinate sum of $E$ is one, the Cartesian equation of a plane isThe distance from a point to a plane is the absolute normal displacement, giving

For the $C$ obtained above, the characteristic polynomial isThus the eigenvalues are $5$ and $-5$. For $\mu =5$, $(C-5I)\mathbf{v}=0$ gives $v_2=-3v_1$; for $\mu =-5$, it gives $v_1=3v_2$. Normalizing the eigenvectors yieldsBoth are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.

The original PDF specifies evaluation at $(1,1,1)$; the local TeX incorrectly gives $(1,1,0)$. Write the scalar field explicitly as $f=c/(x^2+y^2+z^2)$. Its gradient isA directional derivative measures rate per unit distance, so normalize the direction to the unit vector $\mathbf{n}=(1,1,0)/\sqrt{2}$. ThenUsing the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.

The two prescribed directions have dot product zero. Normalize them toTheir cross product is the perpendicular unit vector ${\mathbf{q}}_3=(2,-1,-1{)}^T/\sqrt{6}$. Using these as columns givesThe columns form an orthonormal basis, so $Q^{T}Q=I$ and $Q$ is an orthogonal matrix.

Normalize the direction to the unit vector $\hat{\mathbf{u}}=(1,1,0{)}^T/\sqrt{2}$. The directional derivative is the dot product of this unit vector with the gradient:

The scalar $\mathbf{r}\cdot \hat{\mathbf{u}}$ is the signed scalar projection along the unit vector $\hat{\mathbf{u}}$. Multiplying by that unit vector makes $P_{\hat{\mathbf{u}}}(\mathbf{r})$ the orthogonal projection of $\mathbf{r}$ onto its span.

For the other expression, the vector triple product identity givesThis is the orthogonal projection onto a plane perpendicular to $\hat{\mathbf{u}}$ through the origin. Its dot product with $\hat{\mathbf{u}}$ is zero, while the difference from $\mathbf{r}$ is parallel to $\hat{\mathbf{u}}$.

The unit vector $\hat{\mathbf{a}}$ specifies the direction of the straight line. Reversing its sign, together with the sign of $\mathbf{c}$, leaves the same straight line.

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

Using Einstein notation and the Kronecker delta,The unit vector condition is essential in the last step.

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.