Q.Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

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

Therefore

(The other bisector direction $\hat{OA}+\hat{OB}$ gives $\dfrac{46}{\sqrt{195}}$, which is not among the choices; the listed value corresponds to the direction above.)

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:

Let’s use the cross product method to confirm.

Compute $\overrightarrow{AP} \times \vec{v}$:

Magnitude of the cross product:

Magnitude of $\vec{v}$:

So the distance is:

Both methods agree. The distance is $\frac{1}{\sqrt{6}}$, which is option (A).

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 $\lambda$.

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

Compute:

• $i$: $(3\lambda)(-2) - (-\lambda)(3) = -6\lambda + 3\lambda = -3\lambda$
• $j$: $[(4\lambda)(-2) - (-\lambda)(6)] = (-8\lambda + 6\lambda) = -2\lambda$, then with minus sign: $-(-2\lambda)=2\lambda$
• $k$: $(4\lambda)(3) - (3\lambda)(6) = 12\lambda - 18\lambda = -6\lambda$

So $\vec{d}\times\vec{c} = \lambda(-3\hat{i} + 2\hat{j} - 6\hat{k})$.

Its magnitude:

Given $|\vec{d}\times\vec{c}| = 14$, we have $7|\lambda| = 14 \implies |\lambda| = 2$.

6. Now compute $|\vec{d}\cdot\vec{c}|$ From step 4:

Compute $104 \cdot 49 = 104 \cdot (50-1) = 5200 - 104 = 5096$.

So $|\vec{d}\cdot\vec{c}|^2 = 5096 - 196 = 4900$.

Hence $|\vec{d}\cdot\vec{c}| = \sqrt{4900} = 70$.

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 $k$.

Comparing the components, we get:

4. Use the given condition to find $k$: We are given the condition $\alpha + \beta + \gamma = 46$. Substitute the expressions for $\alpha, \beta, \gamma$ in terms of $k$:

Both 46 and 69 are divisible by 23 ($46 = 2 \times 23$, $69 = 3 \times 23$).

5. Determine the values of $\alpha, \beta, \gamma$: Now substitute the value of $k$ back into the expressions for $\alpha, \beta, \gamma$:

So, the vector is $12\vec{i} + 20\vec{j} + 14\vec{k}$.

6. Calculate the final expression:

We need to find the value of $\alpha - 2\beta + 3\gamma$.

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