RBSE Solutions Class 12 Maths Chapter 13 Vector Exercise 13.1

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.1

 

Question 1. Compute the magnitude of the following vectors :
\( \vec{a} = \widehat{i} + \widehat{j} + \widehat{k} \)
\( \vec{b} = 2\widehat{i} - 7\widehat{j} - 3\widehat{k} \)
\( \vec{c} = \frac{1}{\sqrt{3}}\widehat{i} + \frac{1}{\sqrt{3}}\widehat{j} - \frac{1}{\sqrt{3}}\widehat{k} \)
Answer:
(i) For vector \( \vec{a} = \widehat{i} + \widehat{j} + \widehat{k} \):
The magnitude \( |\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} \)
\( = \sqrt{1+1+1} \)
\( = \sqrt{3} \)
(ii) For vector \( \vec{b} = 2\widehat{i} - 7\widehat{j} - 3\widehat{k} \):
The magnitude \( |\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} \)
\( = \sqrt{4+49+9} \)
\( = \sqrt{62} \)
(iii) For vector \( \vec{c} = \frac{1}{\sqrt{3}}\widehat{i} + \frac{1}{\sqrt{3}}\widehat{j} - \frac{1}{\sqrt{3}}\widehat{k} \):
The magnitude \( |\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2} \)
\( = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} \)
\( = \sqrt{\frac{3}{3}} \)
\( = \sqrt{1} \)
\( = 1 \)
The magnitudes of vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are \( \sqrt{3} \), \( \sqrt{62} \), and 1, respectively. Finding the magnitude of a vector helps us understand its "length" or "strength" without considering its direction.
In simple words: To find the magnitude of a vector, we square each of its number parts, add them up, and then take the square root. This gives us the length of the vector.

🎯 Exam Tip: Remember that magnitude is always a positive value, representing the length of the vector from its starting point to its end point.

 

Question 2. Write two different vectors having same magnitude.
Answer:
Let's consider two different vectors, for example:
\( \vec{a} = 3\widehat{i} + 4\widehat{j} - \widehat{k} \)
\( \vec{b} = 4\widehat{i} - \widehat{j} + 3\widehat{k} \)
Now, we compute their magnitudes:
Magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{3^2 + 4^2 + (-1)^2} \)
\( = \sqrt{9 + 16 + 1} \)
\( = \sqrt{26} \)
Magnitude of \( \vec{b} \):
\( |\vec{b}| = \sqrt{4^2 + (-1)^2 + 3^2} \)
\( = \sqrt{16 + 1 + 9} \)
\( = \sqrt{26} \)
Since both vectors \( \vec{a} \) and \( \vec{b} \) have magnitudes of \( \sqrt{26} \), they have the same magnitude. Vectors can look very different but still have the same overall length, like paths of the same distance in different directions.
In simple words: We can create different vectors by changing the order or signs of their parts. As long as the squared sum of their parts is the same, their total length (magnitude) will be the same.

🎯 Exam Tip: To ensure two vectors have the same magnitude, vary the coefficients of \( \widehat{i} \), \( \widehat{j} \), \( \widehat{k} \) but make sure the sum of their squares remains identical.

 

Question 3. Write two different vectors having same direction.
Answer:
Two vectors have the same direction if one is a positive scalar multiple of the other. This means they point along the same line, just maybe with different lengths.
Let's consider two vectors:
\( \vec{a} = \widehat{i} + \widehat{j} + \widehat{k} \)
\( \vec{b} = 3\widehat{i} + 3\widehat{j} + 3\widehat{k} \)
First, we find the direction cosines (dc's) for vector \( \vec{a} \):
Magnitude of \( \vec{a} \): \( |\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \)
Direction cosines of \( \vec{a} \) are:
\( l_1 = \frac{1}{\sqrt{3}} \)
\( m_1 = \frac{1}{\sqrt{3}} \)
\( n_1 = \frac{1}{\sqrt{3}} \)
Next, we find the direction cosines for vector \( \vec{b} \):
Magnitude of \( \vec{b} \): \( |\vec{b}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \)
Direction cosines of \( \vec{b} \) are:
\( l_2 = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \)
\( m_2 = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \)
\( n_2 = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \)
Since the direction cosines for both vectors \( \vec{a} \) and \( \vec{b} \) are the same \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \), they have the same direction. We can also see that \( \vec{b} = 3\vec{a} \), which confirms they are in the same direction, with \( \vec{b} \) being three times longer than \( \vec{a} \).
In simple words: Two vectors point in the same direction if one vector is just a scaled-up version of the other. You can get one by multiplying the other by a positive number.

🎯 Exam Tip: To check for the same direction, compute the unit vector for each; if they are identical, the directions are the same. Alternatively, verify if one vector is a scalar multiple (positive constant) of the other.

 

Question 4. Find the values of x and y so that the vectors \( 2 \widehat {i} + 3 \widehat {j} \) and \( x \widehat {i } + y \widehat {j} \) are equal.
Answer:
Two vectors are considered equal if their corresponding components are equal. This means the \( \widehat{i} \) components must be the same, and the \( \widehat{j} \) components must be the same.
Given the vectors:
\( \vec{A} = 2\widehat{i} + 3\widehat{j} \)
\( \vec{B} = x\widehat{i} + y\widehat{j} \)
For \( \vec{A} = \vec{B} \), we must have:
The coefficient of \( \widehat{i} \) in \( \vec{A} \) equals the coefficient of \( \widehat{i} \) in \( \vec{B} \)
\( \implies x = 2 \)
The coefficient of \( \widehat{j} \) in \( \vec{A} \) equals the coefficient of \( \widehat{j} \) in \( \vec{B} \)
\( \implies y = 3 \)
So, for the vectors to be equal, the value of \( x \) must be 2 and the value of \( y \) must be 3. This ensures both vectors have the exact same starting point and direction in a 2D plane.
In simple words: If two vectors are equal, it means their 'x' parts are the same, and their 'y' parts are the same. We just compare the numbers in front of \( \widehat{i} \) and \( \widehat{j} \).

🎯 Exam Tip: Remember that for vectors to be equal, all corresponding scalar components (for \( \widehat{i} \), \( \widehat{j} \), and \( \widehat{k} \)) must be exactly the same.

 

Question 5. Find the scalar and vector components of the vector with initial point (2,1) and terminal point (- 5,7).
Answer:
To find the components of a vector given its initial and terminal points, we subtract the coordinates of the initial point from the coordinates of the terminal point.
Initial point A: \( (x_1, y_1) = (2, 1) \)
Terminal point B: \( (x_2, y_2) = (-5, 7) \)
The vector \( \overrightarrow{AB} \) is given by the formula:
\( \overrightarrow{AB} = (x_2 - x_1)\widehat{i} + (y_2 - y_1)\widehat{j} \)
\( \overrightarrow{AB} = (-5 - 2)\widehat{i} + (7 - 1)\widehat{j} \)
\( \overrightarrow{AB} = -7\widehat{i} + 6\widehat{j} \)
From this, we can find the scalar and vector components:
Scalar components are the numbers multiplying the unit vectors. These numbers tell us how much the vector extends along each axis.
Scalar components of \( \overrightarrow{AB} \) are -7 and 6.
Vector components are the scalar components multiplied by their respective unit vectors. These show the actual vector parts along each axis.
Vector components of \( \overrightarrow{AB} \) are \( -7\widehat{i} \) and \( 6\widehat{j} \).
In simple words: To find a vector from one point to another, subtract the starting point's numbers from the ending point's numbers. The plain numbers are scalar components, and the numbers with \( \widehat{i} \), \( \widehat{j} \), \( \widehat{k} \) are vector components.

🎯 Exam Tip: Scalar components are just numbers, while vector components include the direction (like \( \widehat{i} \), \( \widehat{j} \), \( \widehat{k} \)). Pay attention to the signs.

 

Question 6. Find the sum of the vectors \( \vec{a} = \widehat{i} - 2\widehat{j} + \widehat{k} \), \( \vec{b} = -2\widehat{i} + 4\widehat{j} + 5\widehat{k} \) and \( \vec{c} = \widehat{i} - 6\widehat{j} - 7\widehat{k} \).
Answer:
To find the sum of multiple vectors, we add their corresponding scalar components (the numbers in front of \( \widehat{i} \), \( \widehat{j} \), and \( \widehat{k} \)) separately. This is like combining all the 'x' movements, all the 'y' movements, and all the 'z' movements.
Given the vectors:
\( \vec{a} = 1\widehat{i} - 2\widehat{j} + 1\widehat{k} \)
\( \vec{b} = -2\widehat{i} + 4\widehat{j} + 5\widehat{k} \)
\( \vec{c} = 1\widehat{i} - 6\widehat{j} - 7\widehat{k} \)
The sum \( \vec{a} + \vec{b} + \vec{c} \) is:
\( \vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)\widehat{i} + (-2 + 4 - 6)\widehat{j} + (1 + 5 - 7)\widehat{k} \)
\( = (0)\widehat{i} + (-4)\widehat{j} + (-1)\widehat{k} \)
\( = 0\widehat{i} - 4\widehat{j} - \widehat{k} \)
\( = -4\widehat{j} - \widehat{k} \)
Thus, the sum of the given vectors is \( -4\widehat{j} - \widehat{k} \). This resultant vector shows the combined effect of all three individual vectors.
In simple words: To add vectors, just add all the numbers in front of \( \widehat{i} \) together, then all the numbers in front of \( \widehat{j} \) together, and all the numbers in front of \( \widehat{k} \) together. This gives you one new vector.

🎯 Exam Tip: Ensure you correctly combine the coefficients for each unit vector (\( \widehat{i} \), \( \widehat{j} \), \( \widehat{k} \)) and pay close attention to positive and negative signs during addition.

 

Question 7. Find the unit vector in the direction of the vector \( \widehat{a} = \widehat{i} + \widehat{j} + 2\widehat{k} \).
Answer:
A unit vector is a vector that has a magnitude of 1 and points in the same direction as the original vector. To find a unit vector, we divide the original vector by its magnitude.
Given vector \( \vec{a} = \widehat{i} + \widehat{j} + 2\widehat{k} \)
First, find the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} \)
\( = \sqrt{1 + 1 + 4} \)
\( = \sqrt{6} \)
Now, calculate the unit vector \( \widehat{a} \) (sometimes denoted as \( \frac{\vec{a}}{|\vec{a}|} \)):
\( \widehat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\widehat{i} + \widehat{j} + 2\widehat{k}}{\sqrt{6}} \)
\( \widehat{a} = \frac{1}{\sqrt{6}}\widehat{i} + \frac{1}{\sqrt{6}}\widehat{j} + \frac{2}{\sqrt{6}}\widehat{k} \)
This is the required unit vector. This vector is very useful because it only shows the direction, ignoring any 'length' information.
In simple words: To get a unit vector, divide your vector by its total length (magnitude). This makes the new vector exactly one unit long but still pointing in the same way.

🎯 Exam Tip: The unit vector always has a magnitude of 1. If you calculate the magnitude of your resulting unit vector, it should always be 1.

 

Question 8. Find the unit vector in the direction of vector \( \overrightarrow{PQ} \) where P and Q are the points (1,2,3) and (4,5,6).
Answer:
To find the unit vector in the direction of \( \overrightarrow{PQ} \), we first need to find the vector \( \overrightarrow{PQ} \) itself, and then divide it by its magnitude.
Given points:
P \( (x_1, y_1, z_1) = (1, 2, 3) \)
Q \( (x_2, y_2, z_2) = (4, 5, 6) \)
First, find the vector \( \overrightarrow{PQ} \):
\( \overrightarrow{PQ} = (x_2 - x_1)\widehat{i} + (y_2 - y_1)\widehat{j} + (z_2 - z_1)\widehat{k} \)
\( \overrightarrow{PQ} = (4 - 1)\widehat{i} + (5 - 2)\widehat{j} + (6 - 3)\widehat{k} \)
\( \overrightarrow{PQ} = 3\widehat{i} + 3\widehat{j} + 3\widehat{k} \)
Next, find the magnitude of \( \overrightarrow{PQ} \):
\( |\overrightarrow{PQ}| = \sqrt{3^2 + 3^2 + 3^2} \)
\( = \sqrt{9 + 9 + 9} \)
\( = \sqrt{27} \)
\( = 3\sqrt{3} \)
Finally, find the unit vector in the direction of \( \overrightarrow{PQ} \):
\( \widehat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} = \frac{3\widehat{i} + 3\widehat{j} + 3\widehat{k}}{3\sqrt{3}} \)
\( \widehat{PQ} = \frac{3}{3\sqrt{3}}\widehat{i} + \frac{3}{3\sqrt{3}}\widehat{j} + \frac{3}{3\sqrt{3}}\widehat{k} \)
\( \widehat{PQ} = \frac{1}{\sqrt{3}}\widehat{i} + \frac{1}{\sqrt{3}}\widehat{j} + \frac{1}{\sqrt{3}}\widehat{k} \)
This is the unit vector pointing from P to Q, showing only the direction.
In simple words: First, calculate the vector from point P to point Q by subtracting their coordinates. Then, divide this new vector by its length to get a unit vector that only shows direction.

🎯 Exam Tip: Be careful with the order of subtraction for points: terminal point coordinates minus initial point coordinates. Simplify radicals in the magnitude and unit vector for full marks.

 

Question 9. For the given vectors, \( \vec{a} = 2\widehat{i} - \widehat{j} + 2\widehat{k} \) and \( \vec{b} = -\widehat{i} + \widehat{j} - \widehat{k} \). Find the unit vector in the direction of the vector \( \overrightarrow{a} + \overrightarrow{b} \).
Answer:
To find the unit vector in the direction of \( \overrightarrow{a} + \overrightarrow{b} \), we first need to calculate the sum vector \( (\overrightarrow{a} + \overrightarrow{b}) \), and then divide it by its magnitude.
Given vectors:
\( \vec{a} = 2\widehat{i} - \widehat{j} + 2\widehat{k} \)
\( \vec{b} = -\widehat{i} + \widehat{j} - \widehat{k} \)
First, find the sum vector \( \overrightarrow{a} + \overrightarrow{b} \):
\( \overrightarrow{a} + \overrightarrow{b} = (2\widehat{i} - \widehat{j} + 2\widehat{k}) + (-\widehat{i} + \widehat{j} - \widehat{k}) \)
\( = (2 - 1)\widehat{i} + (-1 + 1)\widehat{j} + (2 - 1)\widehat{k} \)
\( = 1\widehat{i} + 0\widehat{j} + 1\widehat{k} \)
\( = \widehat{i} + \widehat{k} \)
Next, find the magnitude of the sum vector \( |\overrightarrow{a} + \overrightarrow{b}| \):
\( |\overrightarrow{i} + \overrightarrow{k}| = \sqrt{1^2 + 0^2 + 1^2} \)
\( = \sqrt{1 + 0 + 1} \)
\( = \sqrt{2} \)
Finally, find the unit vector in the direction of \( (\overrightarrow{a} + \overrightarrow{b}) \):
\( \frac{\overrightarrow{a} + \overrightarrow{b}}{|\overrightarrow{a} + \overrightarrow{b}|} = \frac{\widehat{i} + \widehat{k}}{\sqrt{2}} \)
\( = \frac{1}{\sqrt{2}}\widehat{i} + \frac{1}{\sqrt{2}}\widehat{k} \)
This is the required unit vector. This vector represents the combined direction of both original vectors.
In simple words: First, add the two vectors together by adding their matching parts. Then, find the length of this new sum vector. Finally, divide the sum vector by its length to get the unit vector.

🎯 Exam Tip: Always sum the vectors first before calculating the magnitude for the unit vector. Don't find unit vectors for individual vectors and then sum them.

 

Question 10. Find a vector in the direction of vector \( 5\widehat{i} - \widehat{j} + 2\widehat{k} \) which has magnitude 8 units.
Answer:
To find a vector with a specific magnitude in a given direction, we first find the unit vector in that direction, and then multiply it by the desired magnitude. This allows us to scale a direction-only vector to any length.
Given vector \( \vec{d} = 5\widehat{i} - \widehat{j} + 2\widehat{k} \)
Desired magnitude = 8 units.
First, find the magnitude of \( \vec{d} \):
\( |\vec{d}| = \sqrt{5^2 + (-1)^2 + 2^2} \)
\( = \sqrt{25 + 1 + 4} \)
\( = \sqrt{30} \)
Next, find the unit vector in the direction of \( \vec{d} \):
\( \widehat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{5\widehat{i} - \widehat{j} + 2\widehat{k}}{\sqrt{30}} \)
\( \widehat{d} = \frac{5}{\sqrt{30}}\widehat{i} - \frac{1}{\sqrt{30}}\widehat{j} + \frac{2}{\sqrt{30}}\widehat{k} \)
Finally, multiply this unit vector by the desired magnitude (8) to get the required vector:
Required vector \( = 8 \times \widehat{d} \)
\( = 8 \left( \frac{5}{\sqrt{30}}\widehat{i} - \frac{1}{\sqrt{30}}\widehat{j} + \frac{2}{\sqrt{30}}\widehat{k} \right) \)
\( = \frac{8(5\widehat{i} - \widehat{j} + 2\widehat{k})}{\sqrt{30}} \)
This vector points in the same direction as \( 5\widehat{i} - \widehat{j} + 2\widehat{k} \) but has a length of 8 units.
In simple words: First, turn the given vector into a 'unit' vector (length 1). Then, multiply this unit vector by the new length (8) you want. This makes a new vector that points the same way but has the desired length.

🎯 Exam Tip: A vector in a specific direction with a given magnitude 'k' is found by \( k \times \frac{\vec{a}}{|\vec{a}|} \). Always calculate the unit vector first.

 

Question 11. Show that the vectors \( 2\widehat{i} - 3\widehat{j} + 4\widehat{k} \) and \( -4\widehat{i} + 6\widehat{j} - 8\widehat{k} \) are collinear.
Answer:
Two vectors are collinear if one is a scalar multiple of the other, meaning they lie on the same line or parallel lines. This also includes cases where they point in opposite directions.
Let \( \vec{a} = 2\widehat{i} - 3\widehat{j} + 4\widehat{k} \)
Let \( \vec{b} = -4\widehat{i} + 6\widehat{j} - 8\widehat{k} \)
We need to check if \( \vec{b} = \lambda \vec{a} \) for some scalar \( \lambda \).
Comparing the components:
For \( \widehat{i} \): \( -4 = \lambda (2) \implies \lambda = -2 \)
For \( \widehat{j} \): \( 6 = \lambda (-3) \implies \lambda = -2 \)
For \( \widehat{k} \): \( -8 = \lambda (4) \implies \lambda = -2 \)
Since the value of \( \lambda \) is the same (-2) for all components, vector \( \vec{b} \) can be written as \( -2 \) times vector \( \vec{a} \).
So, \( \vec{b} = -2\vec{a} \).
Since \( \vec{b} \) is a scalar multiple of \( \vec{a} \), the vectors \( \vec{a} \) and \( \vec{b} \) are collinear. They point in opposite directions because \( \lambda \) is negative, but they lie on the same line.
In simple words: Vectors are collinear if you can get one by just multiplying the other by a single number. If that number is the same for all parts of the vector, then they are collinear.

🎯 Exam Tip: Vectors \( \vec{a} \) and \( \vec{b} \) are collinear if \( \vec{b} = \lambda \vec{a} \) where \( \lambda \) is any scalar. A negative \( \lambda \) means they are in opposite directions.

 

Question 12. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \( \widehat{i} + 2\widehat{j} - \widehat{k} \) and \( -\widehat{i} + \widehat{j} + \widehat{k} \) respectively in the ratio 2:1.
(i) internally
(ii) externally
Answer:
To find the position vector of a point R that divides a line segment joining two points P and Q, we use the section formula. The formula changes slightly depending on whether the division is internal or external.
Position vector of P: \( \overrightarrow{OP} = \vec{p} = \widehat{i} + 2\widehat{j} - \widehat{k} \)
Position vector of Q: \( \overrightarrow{OQ} = \vec{q} = -\widehat{i} + \widehat{j} + \widehat{k} \)
Ratio \( m:n = 2:1 \)

(i) Internally:
When point R divides the line segment PQ internally in the ratio \( m:n \), its position vector \( \overrightarrow{OR} \) is given by:
\( \overrightarrow{OR} = \frac{m\vec{q} + n\vec{p}}{m + n} \)
\( \overrightarrow{OR} = \frac{2(-\widehat{i} + \widehat{j} + \widehat{k}) + 1(\widehat{i} + 2\widehat{j} - \widehat{k})}{2 + 1} \)
\( \overrightarrow{OR} = \frac{(-2\widehat{i} + 2\widehat{j} + 2\widehat{k}) + (\widehat{i} + 2\widehat{j} - \widehat{k})}{3} \)
\( \overrightarrow{OR} = \frac{(-2 + 1)\widehat{i} + (2 + 2)\widehat{j} + (2 - 1)\widehat{k}}{3} \)
\( \overrightarrow{OR} = \frac{-1\widehat{i} + 4\widehat{j} + 1\widehat{k}}{3} \)
\( \overrightarrow{OR} = -\frac{1}{3}\widehat{i} + \frac{4}{3}\widehat{j} + \frac{1}{3}\widehat{k} \)

(ii) Externally:
When point R divides the line segment PQ externally in the ratio \( m:n \), its position vector \( \overrightarrow{OR} \) is given by:
\( \overrightarrow{OR} = \frac{m\vec{q} - n\vec{p}}{m - n} \)
\( \overrightarrow{OR} = \frac{2(-\widehat{i} + \widehat{j} + \widehat{k}) - 1(\widehat{i} + 2\widehat{j} - \widehat{k})}{2 - 1} \)
\( \overrightarrow{OR} = \frac{(-2\widehat{i} + 2\widehat{j} + 2\widehat{k}) - (\widehat{i} + 2\widehat{j} - \widehat{k})}{1} \)
\( \overrightarrow{OR} = (-2 - 1)\widehat{i} + (2 - 2)\widehat{j} + (2 - (-1))\widehat{k} \)
\( \overrightarrow{OR} = -3\widehat{i} + 0\widehat{j} + 3\widehat{k} \)
\( \overrightarrow{OR} = -3\widehat{i} + 3\widehat{k} \)
These are the position vectors for internal and external division. The formulas help us locate a point R relative to an origin.
In simple words: To find the position of a point that divides a line, use a special formula. For internal division, you add the cross-multiplied vectors and divide by the sum of ratios. For external division, you subtract them and divide by the difference of ratios.

🎯 Exam Tip: Memorize both the internal (\( \frac{m\vec{q} + n\vec{p}}{m + n} \)) and external (\( \frac{m\vec{q} - n\vec{p}}{m - n} \)) section formulas for position vectors. Pay close attention to signs, especially for external division.

 

Question 13. Find the position vector of the midpoint of the vector joining the points P(2,3,4) and Q(4, 1, -2).
Answer:
The midpoint of a line segment is a special case of internal division where the ratio is 1:1. The position vector of the midpoint is simply the average of the position vectors of the two endpoints. This means the midpoint is exactly halfway between the two points.
Position vector of P: \( \overrightarrow{OP} = \vec{p} = 2\widehat{i} + 3\widehat{j} + 4\widehat{k} \)
Position vector of Q: \( \overrightarrow{OQ} = \vec{q} = 4\widehat{i} + \widehat{j} - 2\widehat{k} \)
The position vector of the midpoint R is given by:
\( \overrightarrow{OR} = \frac{\overrightarrow{OP} + \overrightarrow{OQ}}{2} \)
\( \overrightarrow{OR} = \frac{(2\widehat{i} + 3\widehat{j} + 4\widehat{k}) + (4\widehat{i} + \widehat{j} - 2\widehat{k})}{2} \)
\( \overrightarrow{OR} = \frac{(2 + 4)\widehat{i} + (3 + 1)\widehat{j} + (4 - 2)\widehat{k}}{2} \)
\( \overrightarrow{OR} = \frac{6\widehat{i} + 4\widehat{j} + 2\widehat{k}}{2} \)
\( \overrightarrow{OR} = 3\widehat{i} + 2\widehat{j} + \widehat{k} \)
This is the required position vector of the midpoint. Each component of the midpoint vector is the average of the corresponding components of the two given points.
In simple words: To find the midpoint vector, just add the two given position vectors together, then divide all parts of the new vector by 2. It's like finding the average location.

🎯 Exam Tip: The midpoint formula is a simplified version of the internal division formula where \( m=n=1 \). Ensure you add corresponding components correctly before dividing by 2.

 

Question 14. Show that the points A, B and C with position vectors, \( \vec{a} = 3\widehat{i} - 4\widehat{j} - 4\widehat{k} \), \( \vec{b} = 2\widehat{i} - \widehat{j} + \widehat{k} \) and \( \vec{c} = \widehat{i} - 3\widehat{j} - 5\widehat{k} \) respectively form the vertices of a right angled triangle.
Answer:
To show that three points form the vertices of a right-angled triangle, we can find the vectors representing the sides of the triangle and then use the Pythagorean theorem for magnitudes or the dot product to check for a right angle. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (longest side) equals the sum of the squares of the other two sides. A right angle exists where the dot product of two side vectors is zero.
Let the position vectors of points A, B, and C be:
\( \overrightarrow{OA} = \vec{a} = 3\widehat{i} - 4\widehat{j} - 4\widehat{k} \)
\( \overrightarrow{OB} = \vec{b} = 2\widehat{i} - \widehat{j} + \widehat{k} \)
\( \overrightarrow{OC} = \vec{c} = \widehat{i} - 3\widehat{j} - 5\widehat{k} \)
Now, we find the vectors representing the sides of the triangle:
Vector \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (2\widehat{i} - \widehat{j} + \widehat{k}) - (3\widehat{i} - 4\widehat{j} - 4\widehat{k}) \)
\( = (2 - 3)\widehat{i} + (-1 - (-4))\widehat{j} + (1 - (-4))\widehat{k} \)
\( = -\widehat{i} + 3\widehat{j} + 5\widehat{k} \)
Magnitude of \( \overrightarrow{AB} \):
\( |\overrightarrow{AB}| = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \)
\( |\overrightarrow{AB}|^2 = 35 \)

Vector \( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} \)
\( = (\widehat{i} - 3\widehat{j} - 5\widehat{k}) - (2\widehat{i} - \widehat{j} + \widehat{k}) \)
\( = (1 - 2)\widehat{i} + (-3 - (-1))\widehat{j} + (-5 - 1)\widehat{k} \)
\( = -\widehat{i} - 2\widehat{j} - 6\widehat{k} \)
Magnitude of \( \overrightarrow{BC} \):
\( |\overrightarrow{BC}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41} \)
\( |\overrightarrow{BC}|^2 = 41 \)

Vector \( \overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} \)
\( = (3\widehat{i} - 4\widehat{j} - 4\widehat{k}) - (\widehat{i} - 3\widehat{j} - 5\widehat{k}) \)
\( = (3 - 1)\widehat{i} + (-4 - (-3))\widehat{j} + (-4 - (-5))\widehat{k} \)
\( = 2\widehat{i} - \widehat{j} + \widehat{k} \)
Magnitude of \( \overrightarrow{CA} \):
\( |\overrightarrow{CA}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)
\( |\overrightarrow{CA}|^2 = 6 \)

Now, check the Pythagorean theorem:
We see that \( |\overrightarrow{AB}|^2 = 35 \), \( |\overrightarrow{BC}|^2 = 41 \), and \( |\overrightarrow{CA}|^2 = 6 \).
The sum of the squares of the two shorter sides is:
\( |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = 35 + 6 = 41 \)
This sum is equal to the square of the longest side, \( |\overrightarrow{BC}|^2 = 41 \).
Since \( |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = |\overrightarrow{BC}|^2 \), the points A, B, and C form the vertices of a right-angled triangle, with the right angle at A (opposite to the longest side BC).
In simple words: To show it's a right triangle, first find the vectors for each side. Then, find the length (magnitude) of each side. If the square of the longest side's length equals the sum of the squares of the other two sides' lengths, then it's a right-angled triangle.

🎯 Exam Tip: To prove a right-angled triangle using vectors, calculate the squared magnitudes of all three sides. If the sum of two smaller squared magnitudes equals the largest squared magnitude, it's a right triangle.

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