Q.If either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$, then $\vec{a}\times\vec{b}=\vec{0}$. Is the converse true? Justify your answer with an example.

Let $\vec u = \vec a - \vec b$. Then the second factor is $\vec u - \vec c$, and

Expanding: $-(\vec a\times\vec c) + (\vec b\times\vec c)$. Now dot with $\vec a + 2\vec b - \vec c$ (write $[abc] = \vec a\cdot(\vec b\times\vec c)$). …

A parallelepiped is a three-dimensional figure formed by six parallelograms. Its volume can be understood as the product of the area of its base and its perpendicular height. The problem asks for this height, given the three coterminous edges $\vec{a}, \vec{b}, \vec{c}$ and specifying that $\vec{a}$ and $\vec{b}$ form the base.

The key idea is to use the vector operations that represent these geometric quantities:

1. The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of their scalar triple product, $|[\vec{a} \ \vec{b} \ \vec{c}]|$.
2. The area of the base formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product, $|\vec{a} \times \vec{b}|$.
3. The fundamental relationship between volume, base area, and height is $V = \text{Area of Base} \times \text{Height}$.

We can combine these relationships to find the height.

1. Identify the volume of the parallelepiped: The volume $V$ of the parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of their scalar triple product.

2. Identify the area of the base: The problem states that $\vec{a}$ and $\vec{b}$ determine the base of the parallelepiped. The area of the parallelogram formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product. …

With $\vec{a} = 2\vec{i} - \vec{j} + 2\vec{k}$ and $\vec{b} = 3\vec{i} - 2\vec{j} - 5\vec{k}$.

The "projection vector of $\vec{b}$ on a vector perpendicular to $\vec{a}$" is the component of $\vec{b}$ orthogonal to $\vec{a}$:

Dot product: $\vec{a}\cdot\vec{b} = (2)(3) + (-1)(-2) + (2)(-5) = 6 + 2 - 10 = -2$.

Magnitude squared: $|\vec{a}|^2 = 2^2 + (-1)^2 + 2^2 = 9$.

Projection on $\vec{a}$: …

The problem gives two geometric relationships among four vectors. The first is that $\vec{a}$ is perpendicular only to $\vec{c}$ — this means $\vec{a} \cdot \vec{c} = 0$, but $\vec{a}$ is not perpendicular to $\vec{b}$ or $\vec{d}$ (so $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{a} \cdot \vec{d} \neq 0$ in general). The second is that $\vec{b}$ is parallel to $(\vec{c} - \vec{d})$, which means $\vec{c} - \vec{d} = \lambda \vec{b}$ for some scalar $\lambda$.

Our goal is to find $\vec{c}$ in terms of $\vec{b}$ and $\vec{d}$ alone, eliminating $\lambda$ using the perpendicularity condition.

1. Write the parallelism condition. Since $\vec{b} \parallel (\vec{c} - \vec{d})$, there exists a scalar $\lambda$ such that

Rearranging gives

So $\vec{c}$ is a linear combination of $\vec{d}$ and $\vec{b}$.

2. Use the perpendicularity condition. We know $\vec{a} \perp \vec{c}$, so $\vec{a} \cdot \vec{c} = 0$. Substitute the expression for $\vec{c}$:

This expands to

3. Solve for $\lambda$. Since $\vec{a}$ is not perpendicular to $\vec{b}$ (it is perpendicular only to $\vec{c}$), we have $\vec{a} \cdot \vec{b} \neq 0$. Thus

4. Substitute $\lambda$ back. …

Concept: When are two lines coplanar?

Two lines in 3D space can either be:

• Coplanar: They lie in the same plane (either intersecting or parallel)
• Skew: They don't intersect and aren't parallel

For lines $\vec{r} = \vec{a} + t\vec{b}$ and $\vec{r} = \vec{c} + s\vec{d}$, they are coplanar if and only if the vectors $\vec{b}$ (direction of line 1), $\vec{d}$ (direction of line 2), and $\vec{c} - \vec{a}$ (connecting the two points) are coplanar. This happens when their scalar triple product equals zero.

Solution

1. Find the connecting vector $\vec{c} - \vec{a}$:

2. Compute the cross product $\vec{b} \times \vec{d}$:

Given $\vec{b} = 2\hat{i} - \hat{j} + \lambda\hat{k}$ and $\vec{d} = \hat{i} + 2\hat{j} - \hat{k}$:

3. Compute the scalar triple product:

4. Apply the coplanarity condition:

The lines are coplanar when:

$\lambda = \frac{19}{3}$ …

Expand using bilinearity and $\vec a\times\vec a=\vec b\times\vec b=\vec 0$.

With $\vec a\times\vec a=\vec b\times\vec b=\vec 0$ and $\vec b\times\vec a=-(\vec a\times\vec b)$,

Take the magnitude. …

Interpret the cross-product condition.

so $\vec d-\vec b$ is parallel to $\vec a$, i.e.

Apply $\vec d\cdot\vec c=8$ with $\vec c=2\hat i+\hat j+2\hat k$:

Determine $\vec d$. …

$\vec r$ is perpendicular to both $\vec b$ and $\vec c$, so $\vec r\parallel\vec b\times\vec c$:

$\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-1&5\\1&-4&-2\end{vmatrix}=(22,\ 11,\ -11)=11(2,1,-1).$ …

The first thing to notice is that two lines in 3D can be related in only a few ways: they can be parallel, intersecting, or skew (neither parallel nor intersecting). Perpendicularity is a special case of intersecting lines, but we also check it for direction vectors even if they don't meet. So the plan is simple — compare direction vectors first, then see if the lines intersect.

1. Extract direction vectors.

   For $L_1$, the direction vector is $\vec{d}_1 = 4\vec{i} - 4\vec{j} + 5\vec{k}$.

   For $L_2$, the direction vector is $\vec{d}_2 = 8\vec{i} - 3\vec{j} + \vec{k}$.

2. Check if they are parallel.

   Two lines are parallel if one direction vector is a scalar multiple of the other.

   Is there a scalar $\lambda$ such that $\vec{d}_1 = \lambda \vec{d}_2$?

   Compare components:

   $4 = 8\lambda \implies \lambda = \frac{1}{2}$

   $-4 = -3\lambda \implies \lambda = \frac{4}{3}$

   The two values of $\lambda$ are different, so no single scalar works. Hence the lines are not parallel.

3. Check if they are perpendicular.

   Two lines are perpendicular if their direction vectors are orthogonal, i.e., dot product is zero.

   $\vec{d}_1 \cdot \vec{d}_2 = (4)(8) + (-4)(-3) + (5)(1) = 32 + 12 + 5 = 49 \neq 0$.

   So they are not perpendicular either.

4. Check if they intersect (to decide skew vs. intersecting).

   Write parametric equations.

   For $L_1$:

   $x = 1 + 4t$, $y = 5 - 4t$, $z = 5 + 5t$

   For $L_2$:

   $x = 2 + 8s$, $y = 4 - 3s$, $z = 5 + s$

   For intersection, there must exist $t$ and $s$ satisfying all three equations simultaneously.

   From $x$: $1 + 4t = 2 + 8s \implies 4t - 8s = 1$ …(1) …

Since $\pi$ is parallel to $x-y-z=0$, it shares the normal $(1,-1,-1)$, so $\pi:\ x-y-z=d$.

Substituting the point $(1,-2,1)$:

$1-(-2)-1 = 2 \implies d=2.$ …

We are told that plane $\pi$ passes through $P(3, -4, 5)$ and is parallel to another plane. That other plane passes through $Q(1, 1, -1)$ and is perpendicular to the vector $\vec{n} = \vec{i} + 2\vec{j} - 3\vec{k}$.

Concept & Intuition

If two planes are parallel, they have the same normal vector. So the normal of $\pi$ is exactly the normal of the plane it is parallel to. That second plane is perpendicular to $\vec{n}$, meaning $\vec{n}$ itself is a normal vector to that plane. Therefore $\vec{n}$ is also the normal to $\pi$. Once we have a normal and a point, the equation of $\pi$ is immediate.

1. Identify the normal vector

   The plane through $Q$ is perpendicular to $\vec{i} + 2\vec{j} - 3\vec{k}$. A plane perpendicular to a vector means that vector is normal to the plane. So the normal to that plane is $\vec{n} = (1, 2, -3)$.

   Since $\pi$ is parallel to that plane, $\pi$ has the same normal: $\vec{n} = (1, 2, -3)$.

2. Write the general equation of $\pi$

   A plane with normal $(a, b, c)$ has equation $a x + b y + c z = d$. Here:

3. Use the given point on $\pi$ to find $d$ $\pi$ passes through $(3, -4, 5)$. Substitute: …