Q.Represent graphically a displacement of 40 km, $30^\circ$ east of north.

We are given two lines:

with $\vec{a} = \vec{i} + 2\vec{j}$, $\vec{b} = \vec{i} + \vec{j} + \vec{k}$, $\vec{p} = 2\vec{i} + \vec{k}$, and point $A = 5\vec{i} + \vec{j} - 3\vec{k}$.

Notice that both lines have the same direction vector $\vec{b}$, so they are parallel. The feet of the perpendiculars from A to these lines are points B (on the first line) and C (on the second line). We need $\overrightarrow{BA} + \overrightarrow{AC}$.

Concept and intuition:

Since the lines are parallel, the perpendicular from A to each line will be along the same direction relative to the line’s direction. The vector from A to its foot on a line is simply the component of (A – point on line) perpendicular to $\vec{b}$. The sum $\overrightarrow{BA} + \overrightarrow{AC}$ is actually $\overrightarrow{BA} - \overrightarrow{CA}$, which is the vector from B to C? Let’s see:

$\overrightarrow{BA} = A - B$, $\overrightarrow{AC} = C - A$. Their sum is $(A - B) + (C - A) = C - B$. So we just need the vector from B to C. That’s much simpler: find the feet B and C, then subtract.

Step-by-step:

1. Find foot B on the first line. The first line: $\vec{r} = \vec{a} + t\vec{b}$. Let $B = \vec{a} + t_0 \vec{b}$. The vector $\overrightarrow{AB} = B - A$ must be perpendicular to $\vec{b}$ (since B is the foot). So:

Compute $B - A = (\vec{a} + t_0\vec{b}) - A = (\vec{i} + 2\vec{j}) + t_0(\vec{i}+\vec{j}+\vec{k}) - (5\vec{i}+\vec{j}-3\vec{k})$.

Simplify:

Dot with $\vec{b} = \vec{i}+\vec{j}+\vec{k}$:

So $B = \vec{a} + 0\cdot\vec{b} = \vec{i} + 2\vec{j}$.

2. Find foot C on the second line. Second line: $\vec{r} = \vec{p} + l\vec{b}$. Let $C = \vec{p} + l_0 \vec{b}$. Again, $(C - A) \cdot \vec{b} = 0$. Compute $C - A = (2\vec{i} + \vec{k}) + l_0(\vec{i}+\vec{j}+\vec{k}) - (5\vec{i}+\vec{j}-3\vec{k})$. Simplify:

Dot with $\vec{b}$:

So $C = \vec{p} + 0\cdot\vec{b} = 2\vec{i} + \vec{k}$.

3. Compute $\overrightarrow{BA} + \overrightarrow{AC}$. As noted, this equals $C - B$:

We are given $|\vec{a}| = 5$, $|\vec{b}| = 12$, and $|\vec{a} - \vec{b}| = 13$. Notice that $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This is a classic sign that $\vec{a}$ and $\vec{b}$ are perpendicular. Why? Because for any two vectors,

$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$.

If the left side equals the sum of squares, then the dot product must be zero.

Let’s work through it step by step.

1. Use the given to find the dot product. We have

Substitute the known values:

So $169 = 169 - 2(\vec{a} \cdot \vec{b})$, which gives $2(\vec{a} \cdot \vec{b}) = 0$, hence $\vec{a} \cdot \vec{b} = 0$.

This means $\vec{a}$ and $\vec{b}$ are perpendicular.

2. Now compute $|2\vec{a} + \vec{b}|$. Square it:

Since $\vec{a} \cdot \vec{b} = 0$, the cross term vanishes:

3. Take the square root:

Setup. $L$ passes through $A$ with direction $\vec d = 2\vec i+\vec j-2\vec k$, and $|\vec d|=\sqrt{4+1+4}=3$, so the unit vector is $\hat d=\tfrac{1}{3}(2\vec i+\vec j-2\vec k)$.

Since $P$ lies on $L$ and $|\overrightarrow{AP}|=12$,

Compute the offset. $12\hat d = \tfrac{12}{3}(2,1,-2)=4(2,1,-2)=(8,4,-8)$.

With $\vec P=(-7,-5,11)$:

The first, $-15\vec i-9\vec j+19\vec k$, is a listed choice; $(1,-1,3)$ is not offered.

We are given a line through point A with position vector $\vec{a} = \vec{i} + 2\vec{j} - 2\vec{k}$, parallel to the vector $\vec{v} = 2\vec{i} - 3\vec{j} - 6\vec{k}$. Any point P on this line can be written as:

where $t$ is a real parameter. The distance AP is given as 21, which means the magnitude of the vector $\vec{p} - \vec{a} = t\vec{v}$ is 21.

Why this works: The line is defined by a point and a direction; moving along the direction by a scalar multiple gives all points. The distance condition then fixes the scalar, but since distance is absolute, we get two possible points (one on each side of A). We then match these to the given options.

Let’s work through it step by step.

1. Write the parametric form of P.

So:

2. Express the vector AP.

Its magnitude is:

3. Set the distance equal to 21.

So $t = 3$ or $t = -3$.

4. Find the two possible position vectors for P.
   • For $t = 3$:

• For $t = -3$:

5. Match with the options. The vector $-5\vec{i} + 11\vec{j} + 16\vec{k}$ appears exactly as option (C). The other computed vector $7\vec{i} - 7\vec{j} - 20\vec{k}$ is not listed. However, note that option (A) is $6\vec{i} - 9\vec{j} - 18\vec{k}$. Is that a multiple of the direction vector? Yes: $3 \times (2\vec{i} - 3\vec{j} - 6\vec{k}) = 6\vec{i} - 9\vec{j} - 18\vec{k}$. But that would be the vector from the origin, not from A. Wait — careful: The position vector of P is $\vec{a} + t\vec{v}$. For $t=3$, we got $7\vec{i} - 7\vec{j} - 20\vec{k}$, not $6\vec{i} - 9\vec{j} - 18\vec{k}$. So option (A) is not a point on the line through A? Let’s check: Does there exist a $t$ such that $\vec{a} + t\vec{v} = 6\vec{i} - 9\vec{j} - 18\vec{k}$? Equate components:

These are inconsistent, so (A) is not on the line. Similarly, check (B): $6\vec{i} + 9\vec{j} - 18\vec{k}$ gives $1+2t=6 \Rightarrow t=2.5$, $2-3t=9 \Rightarrow t=-7/3$, no. (D): $5\vec{i} - 11\vec{j} + 16\vec{k}$ gives $1+2t=5 \Rightarrow t=2$, $2-3t=-11 \Rightarrow t=13/3$, no. Only (C) works: $1+2t = -5 \Rightarrow t=-3$, $2-3t=11 \Rightarrow t=-3$, $-2-6t=16 \Rightarrow t=-3$. So (C) is the point for $t=-3$.

Thus the position vector of P can be $-5\vec{i} + 11\vec{j} + 16\vec{k}$.

When you shift the origin to a new point, you are effectively re-centering your coordinate system. The old coordinates $(x, y)$ and the new coordinates $(X, Y)$ are related by a simple translation: if the new origin is at $(h, k)$ in the old system, then $x = X + h$ and $y = Y + k$. Here, the new origin is at $(-1, 2)$, so $h = -1$ and $k = 2$.

The key insight: the given equation $x^2 + y^2 + 2x - 4y + 1 = 0$ actually represents a circle. Completing the square in the old coordinates would reveal its centre at $(-1, 2)$ — exactly the point to which we shift the origin. So the translation should eliminate the linear terms and leave a pure $X^2 + Y^2$ form. Let’s verify this step by step.

1. Write the translation relations. With the new origin at $(-1, 2)$, we have:

2. Substitute into the given equation. The equation is $x^2 + y^2 + 2x - 4y + 1 = 0$. Replace $x$ and $y$:

3. Expand each term carefully.

   • $(X - 1)^2 = X^2 - 2X + 1$
   • $(Y + 2)^2 = Y^2 + 4Y + 4$
   • $2(X - 1) = 2X - 2$
   • $-4(Y + 2) = -4Y - 8$
   • The constant $+1$ remains.

   Now add them all:

4. Combine like terms.

   • $X^2$ and $Y^2$: $X^2 + Y^2$
   • $X$ terms: $-2X + 2X = 0$
   • $Y$ terms: $4Y - 4Y = 0$
   • Constants: $1 + 4 - 2 - 8 + 1 = (1+4) = 5$, then $5 - 2 = 3$, then $3 - 8 = -5$, then $-5 + 1 = -4$

   So we get:

5. Write the transformed equation.

The key idea is that for two skew lines, the shortest distance is the length of the common perpendicular segment between them. This segment is parallel to the cross product of the direction vectors of the two lines. So we need to find the component of the vector joining any point on one line to any point on the other line, along that common perpendicular direction.

Let’s see why each option works or fails.

1. Understanding the given lines.

   The first line is $\vec{r} = \vec{b} + t\vec{a}$ — it passes through point $B$ (position vector $\vec{b}$) and runs along direction $\vec{a}$.

   The second line is $\vec{r} = \vec{d} + s\vec{c}$ — it passes through point $D$ (position vector $\vec{d}$) and runs along direction $\vec{c}$.

2. What is the shortest distance?

   The shortest distance between two skew lines is the length of the line segment that is perpendicular to both lines. This segment is parallel to $\vec{a} \times \vec{c}$ (the cross product of the direction vectors).

   The vector joining a point on the first line to a point on the second line is $\vec{b} - \vec{d}$ (or $\vec{d} - \vec{b}$, sign doesn’t matter for magnitude).

   The shortest distance is the magnitude of the projection of this joining vector onto the unit vector along $\vec{a} \times \vec{c}$.

   $\text{Shortest distance} = \left| \frac{(\vec{b} - \vec{d}) \cdot (\vec{a} \times \vec{c})}{|\vec{a} \times \vec{c}|} \right|$

   This is exactly the orthogonal projection of $(\vec{b} - \vec{d})$ onto $(\vec{a} \times \vec{c})$.

3. Checking the options.

   • (A) Magnitude of $\vec{b} \times \vec{d}$: This has nothing to do with the direction vectors $\vec{a}$ and $\vec{c}$ — it’s just the area of a parallelogram formed by the two position vectors. Wrong.
   • (B) Sum of projections of $\vec{b}$ on $\vec{d}$ and $\vec{d}$ on $\vec{b}$: This is a scalar sum unrelated to the perpendicular distance between the lines. Wrong.
   • (C) Orthogonal projection of $(\vec{a} - \vec{c})$ on $(\vec{b} \times \vec{d})$: Here the vector being projected is the difference of direction vectors, not the vector joining the points. And the direction of projection uses the cross product of position vectors, not direction vectors. Wrong.
   • (D) Orthogonal projection of $(\vec{b} - \vec{d})$ on $(\vec{a} \times \vec{c})$: This matches exactly the formula above. Correct.

The core concept here is the triangle law of vector addition. This law states that if two vectors are represented by two sides of a triangle taken in order, then their resultant is represented by the third side taken in the opposite order.

In simpler terms, if you go from point A to B (vector $\overrightarrow{AB}$) and then from B to C (vector $\overrightarrow{BC}$), the net displacement is from A to C (vector $\overrightarrow{AC}$). So, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.

This principle can be extended: if you have a series of vectors where the terminal point of one vector is the initial point of the next, their sum is a vector from the initial point of the first to the terminal point of the last. For example, $\overrightarrow{P_1P_2} + \overrightarrow{P_2P_3} + \dots + \overrightarrow{P_{n-1}P_n} = \overrightarrow{P_1P_n}$.

We can rearrange the given vectors and group them in a way that allows us to apply this law repeatedly to simplify the expression. Remember that vector addition is commutative and associative, meaning the order of addition doesn't matter, and we can group vectors as we wish.

1. List the given vectors:

   We need to find the resultant of the following six vectors:

   $\overrightarrow{AB}$, $\overrightarrow{AE}$, $\overrightarrow{BC}$, $\overrightarrow{DC}$, $\overrightarrow{ED}$, $\overrightarrow{AC}$

2. Group vectors for initial simplification:

   Let the resultant vector be $\vec{R}$.

   $\vec{R} = \overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC}$

   We can rearrange the terms to group vectors that form a chain:

   $\vec{R} = (\overrightarrow{AB} + \overrightarrow{BC}) + (\overrightarrow{ED} + \overrightarrow{DC}) + \overrightarrow{AE} + \overrightarrow{AC}$

3. Apply the triangle law to the first group:

   Consider the group $(\overrightarrow{AB} + \overrightarrow{BC})$.

   $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$

   Applying this, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.

4. Apply the triangle law to the second group:

   Consider the group $(\overrightarrow{ED} + \overrightarrow{DC})$.

   Applying the triangle law, $\overrightarrow{ED} + \overrightarrow{DC} = \overrightarrow{EC}$.

5. Substitute the simplified groups back into the resultant expression:

   Now, substitute the results from steps 3 and 4 back into the expression for $\vec{R}$:

   $\vec{R} = \overrightarrow{AC} + \overrightarrow{EC} + \overrightarrow{AE} + \overrightarrow{AC}$

6. Rearrange and apply the triangle law again:

   We can rearrange the terms to form another chain:

   $\vec{R} = \overrightarrow{AC} + (\overrightarrow{AE} + \overrightarrow{EC}) + \overrightarrow{AC}$

   Now, consider the group $(\overrightarrow{AE} + \overrightarrow{EC})$.

   Applying the triangle law again, $\overrightarrow{AE} + \overrightarrow{EC} = \overrightarrow{AC}$.

7. Final simplification:

   Substitute this back into the expression for $\vec{R}$:

   $\vec{R} = \overrightarrow{AC} + \overrightarrow{AC} + \overrightarrow{AC}$

   $\vec{R} = 3\overrightarrow{AC}$

The resultant vector is $3\overrightarrow{AC}$.