Q.If with reference to the right handed system of mutually perpendicular unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$, $\vec{\alpha}=3\hat{i}-\hat{j}$ and $\vec{\beta}=2\hat{i}+\hat{j}-3\hat{k}$, then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}$, where $\vec{\beta_1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta_2}$ is perpendicular to $\vec{\alpha}$.

The core idea here is that the cross product of two non-parallel vectors yields a third vector that is perpendicular to both of the original vectors. If two vectors lie in a plane, any vector perpendicular to both of them must also be perpendicular to the plane containing them.

1. Identify the vectors defining the plane:

   We are given two vectors, $\vec{a} = \vec{i} - 2\vec{j} + 2\vec{k}$ and $\vec{b} = 2\vec{i} + 3\vec{j} - 6\vec{k}$. The plane in question is defined by the vectors $2\vec{a} + \vec{b}$ and $\vec{b} - \vec{a}$. Let's calculate these two vectors.

   First, calculate $2\vec{a} + \vec{b}$:

Let $\vec{u} = 4\vec{i} - \vec{j} - 2\vec{k}$.

Next, calculate $\vec{b} - \vec{a}$:

Let $\vec{v} = \vec{i} + 5\vec{j} - 8\vec{k}$.

2. Find a vector perpendicular to the plane:

A vector perpendicular to the plane containing $\vec{u}$ and $\vec{v}$ is given by their cross product, $\vec{u} \times \vec{v}$.

> [!FORMULA]

> The cross product of two vectors $\vec{A} = A_x\vec{i} + A_y\vec{j} + A_z\vec{k}$ and $\vec{B} = B_x\vec{i} + B_y\vec{j} + B_z\vec{k}$ is given by:

> $\vec{A} \times \vec{B} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$

Let $\vec{n} = \vec{u} \times \vec{v}$:

This vector $\vec{n}$ is perpendicular to the plane containing $\vec{u}$ and $\vec{v}$.

3. Relate the given perpendicular vector to $\vec{n}$:

The problem states that $\alpha \vec{i} + \beta \vec{j} + \gamma \vec{k}$ is a vector perpendicular to the plane. This means it must be parallel to $\vec{n}$. Therefore, it must be a scalar multiple of $\vec{n}$.

Let $\alpha \vec{i} + \beta \vec{j} + \gamma \vec{k} = k\vec{n}$ for some scalar $k$. …

Here $\overrightarrow{OA}=\hat i+2\hat j+2\hat k$ with $|\overrightarrow{OA}|=3$, and $\overrightarrow{OB}=3\hat i+4\hat k$ with $|\overrightarrow{OB}|=5$, so

The angle bisector at $O$ lies along $\hat{OA}\pm\hat{OB}$. Taking the direction that matches the given options,

A vector along this bisector of length $2$ is …

Concept & Intuition

The problem gives three vectors and says $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. That means $\vec{d}$ is parallel to the cross product $\vec{a}\times\vec{b}$ — because the cross product of two vectors is perpendicular to both. So $\vec{d} = \lambda (\vec{a}\times\vec{b})$ for some scalar $\lambda$.

Then we have a condition involving $\vec{c}$: $|\vec{d}\times\vec{c}| = 14$. Since $\vec{d}$ is parallel to $\vec{a}\times\vec{b}$, the cross product $\vec{d}\times\vec{c}$ will be perpendicular to both $\vec{d}$ and $\vec{c}$. Its magnitude relates to the area of the parallelogram spanned by $\vec{d}$ and $\vec{c}$.

We are asked for $|\vec{d}\cdot\vec{c}|$, which is the absolute value of the scalar projection of $\vec{c}$ onto $\vec{d}$ times the length of $\vec{d}$. There is a neat identity linking $|\vec{d}\times\vec{c}|$ and $|\vec{d}\cdot\vec{c}|$: for any two vectors, $|\vec{d}\times\vec{c}|^2 + |\vec{d}\cdot\vec{c}|^2 = |\vec{d}|^2 |\vec{c}|^2$. This is the vector form of the Pythagorean theorem — it comes from $|\vec{d}\times\vec{c}| = |\vec{d}||\vec{c}|\sin\theta$ and $|\vec{d}\cdot\vec{c}| = |\vec{d}||\vec{c}|\cos\theta$.

So if we can find $|\vec{d}|$ and $|\vec{c}|$, we can get $|\vec{d}\cdot\vec{c}|$ directly.

Step-by-step solution

1. Find $\vec{a}\times\vec{b}$

Compute each:

• $i$ component: $2 - (-2) = 4$
• $j$ component: $(-2 - 1) = -3$, but with minus sign: $-(-3)=3$
• $k$ component: $(-2) - (-1) = -1$

So $\vec{a}\times\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}$.

2. Write $\vec{d}$ as a scalar multiple

   Since $\vec{d} \perp \vec{a}$ and $\vec{d} \perp \vec{b}$, $\vec{d}$ is parallel to $\vec{a}\times\vec{b}$.

   Let $\vec{d} = \lambda (4\hat{i} + 3\hat{j} - \hat{k})$.

3. Use the condition $|\vec{d}\times\vec{c}| = 14$

   First, note that $\vec{d}\times\vec{c} = \lambda (\vec{a}\times\vec{b})\times\vec{c}$.

   But we can also use the magnitude relation:

However, it's easier to compute $|\vec{d}|$ and $|\vec{c}|$ and then use the Pythagorean identity.

Compute $|\vec{c}|$: $\vec{c} = (6,3,-2)$, so

Compute $|\vec{a}\times\vec{b}|$:

Hence $|\vec{d}| = |\lambda| \sqrt{26}$.

4. Apply the Pythagorean identity For any two vectors $\vec{u}, \vec{v}$:

Here $\vec{u}=\vec{d}$, $\vec{v}=\vec{c}$.

We know $|\vec{d}\times\vec{c}| = 14$, so $|\vec{d}\times\vec{c}|^2 = 196$.

Also $|\vec{c}|^2 = 49$, and $|\vec{d}|^2 = \lambda^2 \cdot 26$.

So:

5. Find $\lambda^2$ using another relation

   We haven't used the fact that $\vec{d}\times\vec{c}$ has magnitude 14 directly in terms of $\lambda$. Let's compute $\vec{d}\times\vec{c}$ explicitly to find $\lambda$.

   $\vec{d} = \lambda(4,3,-1)$, $\vec{c} = (6,3,-2)$.

   $$\vec{d}\times\vec{c} = …$$

The key idea is that the shortest distance from a point to a line is the length of the perpendicular segment. In 3D, we can find this by taking the vector from a known point on the line to our point, and then subtracting the component of that vector that lies along the line’s direction. What remains is the perpendicular component, and its magnitude is the distance.

Let’s work through it step by step.

1. Identify the given information.

   The line $L$ passes through point $A$ with position vector $2\vec{i} + \vec{j}$, so $A = (2, 1, 0)$.

   It is parallel to $\vec{v} = \vec{i} + 5\vec{j} + 2\vec{k}$, so the direction vector is $(1, 5, 2)$.

   The given point is $P = (3, 5, 2)$.

2. Find the vector from a point on the line to $P$.

   Take $\overrightarrow{AP} = P - A = (3-2, 5-1, 2-0) = (1, 4, 2)$.

3. Project $\overrightarrow{AP}$ onto the direction vector $\vec{v}$.

   The projection formula gives the component of $\overrightarrow{AP}$ along $\vec{v}$:

Compute the dot products:

So the projection vector is:

4. Find the perpendicular component. The vector from $A$ to $P$ can be split into two parts: one along the line and one perpendicular. The perpendicular component is:

Simplify: $\left( \frac{1}{6}, -\frac{1}{6}, \frac{1}{3} \right)$.

5. Compute the magnitude of the perpendicular component — this is the distance.

Note that $\frac{1}{9} = \frac{4}{36}$, so:

$d = \sqrt{ \frac{1 + 1 + 4}{36} } = \sqrt{ \frac{6}{36} } = \sqrt{ \frac{1}{6} } = \frac{1}{\sqrt{6}}$ …

We are told $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. That means $\vec{d}$ is parallel to the cross product $\vec{a}\times\vec{b}$. So the direction of $\vec{d}$ is fixed; only its magnitude is unknown. The condition $|\vec{d}\times\vec{c}| = 14$ will determine that magnitude, and then $|\vec{d}\cdot\vec{c}|$ follows directly.

1. Find a vector parallel to $\vec{d}$. Compute $\vec{a}\times\vec{b}$:

So $\vec{a}\times\vec{b} = 4\mathbf{i} + 3\mathbf{j} - \mathbf{k}$.

2. Express $\vec{d}$ as a scalar multiple. Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, it must be parallel to $\vec{a}\times\vec{b}$. Hence

for some scalar $\lambda$ (which could be positive or negative; magnitude will be determined).

3. Use the condition $|\vec{d}\times\vec{c}| = 14$. First compute $\vec{d}\times\vec{c}$:

Compute the cross product of the direction vectors:

Therefore

4. Find $|\lambda|$ from the given magnitude. The magnitude is

We are told this equals $14$, so

$7|\lambda| = 14 \quad\Rightarrow\quad |\lambda| = 2.$ …

The area of a triangle with two sides given by vectors $\vec{a}$ and $\vec{b}$ is not simply the product of their lengths — that would give the area of a rectangle. Instead, the triangle's area is half the area of the parallelogram spanned by the two vectors. And the area of that parallelogram is exactly the magnitude of the cross product $|\vec{a} \times \vec{b}|$.

So the key formula is:

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. This works because $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ gives the parallelogram area directly.

Let’s apply it step by step.

1. Find $|\vec{a}|$. $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$, so

2. We are given $|\vec{b}| = 6$ and $\theta = \frac{\pi}{6}$.

   Recall $\sin\frac{\pi}{6} = \frac{1}{2}$.

3. Compute $|\vec{a} \times \vec{b}|$: …

The problem asks for the length of the perpendicular from a given point P to a plane. This length is precisely the shortest distance from the point P to the plane.

Concept and Intuition

The distance from a point to a plane is a fundamental concept in 3D geometry. Imagine a point P and a plane. If you drop a perpendicular from P to the plane, it meets the plane at a point M, which is called the foot of the perpendicular. The length of the segment PM is the shortest distance from P to the plane.

To find this distance, we first need the equation of the plane. A plane is uniquely defined by a point it passes through and a vector perpendicular to it (its normal vector). Once we have the plane's equation in the standard form $Ax + By + Cz + D = 0$, we can use a direct formula.

This formula arises from vector projection. If we take any point $A(x_1, y_1, z_1)$ on the plane, the vector $\vec{AP} = (x_0-x_1)\vec{i} + (y_0-y_1)\vec{j} + (z_0-z_1)\vec{k}$ connects a point on the plane to the given point P. The normal vector to the plane is $\vec{n} = A\vec{i} + B\vec{j} + C\vec{k}$. The perpendicular distance PM is the magnitude of the projection of $\vec{AP}$ onto the normal vector $\vec{n}$.

Expanding this expression leads directly to the formula above.

Step-by-step Derivation

1. Determine the equation of the plane. The plane passes through point $A(3, -2, 1)$ and is perpendicular to the vector $\vec{n} = 4\vec{i} + 7\vec{j} - 4\vec{k}$. The general equation of a plane passing through a point $(x_1, y_1, z_1)$ with a normal vector $A\vec{i} + B\vec{j} + C\vec{k}$ is $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. Substituting the given values: $4(x-3) + 7(y-(-2)) - 4(z-1) = 0$ …

The foot of the perpendicular $F=(2,0,-1)$ from $P=(1,0,-2)$ lies on the plane, and $\vec{PF}$ is along the plane's normal:

$\vec{PF}=(2-1,\,0-0,\,-1-(-2))=(1,0,1).$ …