Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.6 | Set 1

In Class 12 Mathematics, vectors play a crucial role in understanding spatial relationships and solving geometric problems. Chapter 23 of RD Sharma’s textbook "Algebra of Vectors" covers the fundamental concepts of vector algebra which are essential for analyzing vector quantities in the various dimensions. Exercise 23.6 | Set 1 focuses on applying vector algebra concepts to solve practical problems.

Algebra of Vectors

The algebra of vectors involves operations such as addition, subtraction, and scalar multiplication. The Vectors are quantities that have both the magnitude and direction and their algebraic manipulation follows specific rules. Key operations include:

• Vector Addition: Combining two vectors to obtain a resultant vector.
• Vector Subtraction: Finding the difference between the two vectors.
• Scalar Multiplication: The Scaling of a vector by the scalar quantity.

Question 1: Find the magnitude of the vector \vec{a} = 2\hat{i}+3\hat{j}-6\hat{k}  .

Solution:

Magnitude of a vector x\hat{i}+y\hat{j}+z\hat{k} = \sqrt{x^2+y^2+z^2}

=> |\vec{a}| = \sqrt{2^2+3^2+(-6)^2}

=> |\vec{a}| = \sqrt{4+9+36}

=> |\vec{a}| = \sqrt{49}

=> |\vec{a}| = 7

Question 2: Find the unit vector in the direction of 3\hat{i}+4\hat{j}-12\hat{k}  .

Solution:

We know that unit vector of a vector \vec{a}   is given by,

=> \hat{p} = \dfrac{\vec{a}}{|\vec{a}|}

=> \hat{p} = \dfrac{1}{\sqrt{3^2+4^2+(-12)^2}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{9+16+144}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{169}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{13}(3\hat{i}+4\hat{j}-12\hat{k})

Question 3: Find a unit vector in the direction of the resultant of the vectors \hat{i}-\hat{j}+3\hat{k}, 2\hat{i}+\hat{j}-2\hat{k}   and \hat{i}+2\hat{j}-2\hat{k}.

Solution:

Let,

=> \vec{a} = \hat{i}-\hat{j}+3\hat{k}

=> \vec{b} = 2\hat{i}+\hat{j}-2\hat{k}

=> \vec{c} = \hat{i}+2\hat{j}-2\hat{k}

Let \vec{d}   be the resultant,

=> \vec{d} = \vec{a} + \vec{b} + \vec{c}

=> \vec{d} = (\hat{i}-\hat{j}+3\hat{k})+(2\hat{i}+\hat{j}-2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k})

=> \vec{d} = 4\hat{i}+2\hat{j}-\hat{k}

Unit vector is,

=> \hat{p} = \dfrac{\vec{d}}{|\vec{d}|}

=> \hat{p} = \dfrac{1}{\sqrt{4^2+2^2+(-1)^2}}(4\hat{i}+2\hat{j}-\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{16+4+1}}(4\hat{i}+2\hat{j}-\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}+2\hat{j}-\hat{k})

Question 4: The adjacent sides of a parallelogram are represented by the vectors \vec{a}=\hat{i}+\hat{j}-\hat{k}   and \vec{b}=-2\hat{i}+\hat{j}+2\hat{k}. Find the unit vectors parallel to the diagonals of the parallelogram.

Solution:

Let PQRS be the parallelogram.

Given that, PQ = \hat{i}+\hat{j}-\hat{k} and QR = -2\hat{i}+\hat{j}+2\hat{k}.

Thus, the diagonals are: PR and SQ.

=> \vec{PR} = \vec{PQ} + \vec{QR}

=> \vec{PR} = \vec{a} + \vec{b}

=> \vec{PR} = (\hat{i}+\hat{j}-\hat{k})+(-2\hat{i}+\hat{j}+2\hat{k})

=> \vec{PR} = -\hat{i} + 2\hat{j} +\hat{k}

=> \vec{SQ} = \vec{PQ} - \vec{PS}

=> \vec{SQ} = \vec{a} - \vec{b}

=> \vec{SQ} =  (\hat{i}+\hat{j}-\hat{k})-(-2\hat{i}+\hat{j}+2\hat{k})

=> \vec{SQ} = 3\hat{i}-3\hat{k}

Thus the unit vectors in the direction of the diagonals are:

=> \hat{PR} = \dfrac{\vec{PR}}{|\vec{PR}|}

=> \hat{PR} = \dfrac{1}{\sqrt{(-1)^2+2^2+1^2}}( -\hat{i} + 2\hat{j} +\hat{k})

=> \hat{PR} = \dfrac{1}{\sqrt{6}}(-\hat{i}+2\hat{j}+\hat{k})

=> \hat{SQ} = \dfrac{\vec{SQ}}{|\vec{SQ}|}

=> \hat{SQ} = \dfrac{1}{\sqrt{3^2+(-3)^2}}( 3\hat{i}-3\hat{k})

=> \hat{SQ} = \dfrac{1}{3\sqrt{2}}(3\hat{i}-3\hat{k})

Question 5: If \vec{a} = 3\hat{i}-\hat{j}-4\hat{k}, \vec{b}= -2\hat{i}+4\hat{j}-3\hat{k}   and \vec{c}=\hat{i}+2\hat{j}-\hat{k}, find |3\vec{a}-2\vec{b}+4\vec{c}|.

Solution:

Given, \vec{a} = 3\hat{i}-\hat{j}-4\hat{k}, \vec{b}= -2\hat{i}+4\hat{j}-3\hat{k} and \vec{c}=\hat{i}+2\hat{j}-\hat{k}.

Let,

=> \vec{d} = 3\vec{a}-2\vec{b}+4\vec{c}

=> \vec{d} = 3(3\hat{i}-\hat{j}-4\hat{k})-2(-2\hat{i}+4\hat{j}-3\hat{k})+4(\hat{i}+2\hat{j}-\hat{k})

=> \vec{d} = (9\hat{i}-3\hat{j}-12\hat{k})+ (4\hat{i}-8\hat{j}+6\hat{k})+(4\hat{i}+8\hat{j}-4\hat{k})

=> \vec{d} = 17\hat{i}-3\hat{j}-10\hat{k}

The magnitude is given by,

=> |\vec{d}| = \sqrt{17^2+(-3)^2+(-10)^2}

=> |\vec{d}| = \sqrt{289+9+100}

=> |\vec{d}| = \sqrt{398}

Question 6: If \vec{PQ} = 3\hat{i}+2\hat{j}-\hat{k}   and the coordinates of P are (1,-1,2), find the coordinates of Q.

Solution:

Given, \vec{PQ} = 3\hat{i}+2\hat{j}-\hat{k}

And, \vec{P} = \hat{i}-\hat{j}+2\hat{k}

=> \vec{PQ} = \vec{Q}-\vec{P}

=> \vec{Q} = \vec{PQ}+ \vec{P}

=> \vec{Q} = (3\hat{i}+2\hat{j}-\hat{k})+(\hat{i}-\hat{j}+2\hat{k})

=> \vec{Q} = 4\hat{i}+\hat{j}+\hat{k}

=> Thus the coordinates of Q are (4,1,1).

Question 7: Prove that the points \hat{i}-\hat{j}, 4\hat{i}-3\hat{j}+\hat{k} and 2\hat{i}-4\hat{j}+5\hat{k} are the vertices of a right-angled triangle.

Solution:

Let,

=> \vec{A} = \hat{i}-\hat{j}

=> \vec{B} = 4\hat{i}-3\hat{j}+\hat{k}

=> \vec{C} = 2\hat{i}-4\hat{j}+5\hat{k}

Thus, the 3 sides of the triangle are,

=> \vec{AB} = \vec{B} - \vec{A}

=> \vec{AB} = (4\hat{i}-3\hat{j}+\hat{k})-(\hat{i}-\hat{j})

=> \vec{AB} = 3\hat{i}-2\hat{j}+\hat{k}

=> \vec{BC} = \vec{C} - \vec{B}

=> \vec{BC} = (2\hat{i}-4\hat{j}+5\hat{k})-(4\hat{i}-3\hat{j}+\hat{k})

=> \vec{BC} = -2\hat{i}-\hat{j}+4\hat{k}

=> \vec{CA} = \vec{A} -\vec{C}

=> \vec{CA} = ( \hat{i}-\hat{j})-(2\hat{i}-4\hat{j}+5\hat{k})

=> \vec{CA} = -\hat{i}+3\hat{j}-5\hat{k}

The lengths of every side are given by their magnitude,

=> |\vec{AB}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{14}

=> |\vec{BC}| = \sqrt{(-2)^2+(-1)^2+4^2} = \sqrt{21}

=> |\vec{CA}| = \sqrt{(-1)^2+3^2+(-5)^2} = \sqrt{35}

As we can see,

=> |\vec{CA}|^2 = |\vec{AB}|^2+|\vec{BC}|^2

=> These 3 points form a right-angled triangle.

Question 8: If the vertices A, B and C of a triangle ABC are the points with position vectors a_1\hat{i}+a_2\hat{j}+a_3\hat{k}  , b_1\hat{i}+b_2\hat{j}+b_3\hat{k}  , c_1\hat{i}+c_2\hat{j}+c_3\hat{k}   respectively, what are the vectors determined by its sides? Find the length of these vectors.

Solution:

Let,

=> \vec{a} =a_1\hat{i}+a_2\hat{j}+a_3\hat{k}

=> \vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

=> \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}

The sides of the triangle are given as,

=> \vec{AB} = \vec{b} - \vec{a}

=> \vec{AB} = ( b_1\hat{i}+b_2\hat{j}+b_3\hat{k})-(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})

=> \vec{AB} = (b_1-a_1)\hat{i}+(b_2-a_2)\hat{j}+(b_3-a_3)\hat{k}

=> \vec{BC} = \vec{c}-\vec{b}

=> \vec{BC} = ( c_1\hat{i}+c_2\hat{j}+c_3\hat{k})-(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})

=> \vec{BC} = (c_1-b_1)\hat{i}+(c_2-b_2)\hat{j}+(c_3-b_3)\hat{k}

=> \vec{CA} = \vec{a}-\vec{c}

=> \vec{CA} = ( a_1\hat{i}+a_2\hat{j}+a_3\hat{k})-(c_1\hat{i}+c_2\hat{j}+c_3\hat{k})

=> \vec{CA} = (a_1-c_1)\hat{i}+(a_2-c_2)\hat{j}+(a_3-c_3)\hat{k}

The lengths of the sides are,

=> |\vec{AB}| = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}

=> |\vec{BC}| = \sqrt{(c_1-b_1)^2+(c_2-b_2)^2+(c_3-b_3)^2}

=> |\vec{CA}| = \sqrt{(a_1-c_1)^2+(a_2-c_2)^2+(a_3-c_3)^2}

Question 9: Find the vector from the origin O to the centroid of the triangle whose vertices are (1,-1,2), (2,1,3), and (-1,2,-1).

Solution:

The position of the centroid is given by,

=> (x, y, z) = (\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})

=> (x, y, z) = (\dfrac{1+2+(-1)}{3},\dfrac{(-1)+1+2}{3},\dfrac{2+3+(-1)}{3})

=> (x, y, z) = (\dfrac{2}{3},\dfrac{2}{3},\dfrac{4}{3})

The vector to the centroid from O is,

=> \vec{c} = \dfrac{2}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{4}{3}\hat{k}

Question 10: Find the position vector of a point R which divides the line segment joining points p(\hat{i}+2\hat{j}+\hat{k}  ) and q(-\hat{i}+\hat{j}+\hat{k}) in the ratio 2:1.

(i) Internally

Solution:

The position vectors of a point that divides a line segment internally are given by,

=> \vec{OR} = \dfrac{m\vec{Q}+n\vec{P}}{m+n}  , where \dfrac{m}{n}=\dfrac{2}{1}

=> \vec{OR} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2\hat{j}+\hat{k})}{2+1}

=> \vec{OR} = \dfrac{-\hat{i}+4\hat{j}+3\hat{k}}{3}

(ii) Externally

Solution:

The position vectors of a point that divides a line segment externally are given by,

=> \vec{OR} = \dfrac{m\vec{Q}-n\vec{P}}{m-n}  , where \dfrac{m}{n}=\dfrac{2}{1}

=> \vec{OR} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2\hat{j}+\hat{k})}{2-1}

=> \vec{OR} = 3\hat{i}-\hat{k}

Conclusion

Understanding the algebra of vectors is essential for the tackling complex mathematical and physical problems. Exercise 23.6 | Set 1 from RD Sharma provides practice in applying these concepts to solve real-world problems. Mastery of vector algebra not only aids in the academic pursuits but also in the practical applications in physics and engineering.