Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.6 | Set 2

Question 11: Find the position vector of the mid-point of the vector joining the points P(2\hat{i}-3\hat{j}+4\hat{k} ) and Q(4\hat{i}+\hat{j}-2\hat{k} ).

Solution:

The mid-point of the line segment joining 2 vectors is given by:

=> \vec{R} = \dfrac{\vec{P}+\vec{Q}}{2}

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

=> \vec{R} = \dfrac{6\hat{i}-2\hat{j}+2\hat{k}}{2}

=> \vec{R} = 3\hat{i}-\hat{j}+\hat{k}

Question 12: Find the unit vector in the direction of the vector \vec{PQ}, where P and Q are the points (1,2,3) and (4,5,6).

Solution:

Let,

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

=> \vec{q} = 4\hat{i} + 5\hat{j}+6\hat{k}

=> \vec{PQ} = \vec{q}-\vec{p}

=> \vec{PQ} = (4\hat{i} + 5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})

=> \vec{PQ} = 3\hat{i}+3\hat{j}+3\hat{k}

Unit vector is,

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

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

=> \hat{PQ} = \dfrac{1}{3\sqrt{3}}3(\hat{i}+\hat{j}+\hat{k})

=> \hat{PQ} = \dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})

Question 13: Show that the points A(2\hat{i}-\hat{j}+\hat{k}), B(\hat{i}-3\hat{j}-5\hat{k}), C(3\hat{i}-4\hat{j}-4\hat{k}) are the vertices of a right-angled triangle.

Solution:

Let,

=> \vec{a} = 2\hat{i}-\hat{j}+\hat{k}

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

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

The line segments are,

=> \vec{AB} = \vec{b}-\vec{a}

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

=> \vec{AB} = -\hat{i}-2\hat{j}-6\hat{k}

=> \vec{BC} = \vec{c}-\vec{b}

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

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

=> \vec{CA} = \vec{a}-\vec{c}

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

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

The magnitudes of the sides are,

=> |\vec{AB}| = \sqrt{(-1)^2+(-2)^2+(-6)^2} = \sqrt{41}

=> |\vec{BC}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{6}

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

As we can see that |\vec{AB}|^2 = |\vec {BC}|^2+|\vec{CA}|^2

=> Thus, ABC is a right-angled triangle.

Question 14: Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).

Solution:

Let,

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

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

 The mid-point of the line segment joining 2 vectors is given by:

=> \vec{r} = \dfrac{\vec{p}+\vec{q}}{2}

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

=> \vec{r} = \dfrac{6\hat{i}+4\hat{j}+2\hat{k}}{2}

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

Question 15: Find the value of x for which x(\hat{i}+\hat{j}+\hat{k}) is a unit vector.

Solution:

The magnitude of the given vector is,

=> |x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{x^2+x^2+x^2}

=> |x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{3x^2}

=> |x(\hat{i}+\hat{j}+\hat{k})| = \pm x\sqrt{3}

For it to be a unit vector,

=> |x(\hat{i}+\hat{j}+\hat{k})| = 1

=> x\sqrt{3} = \pm 1

=> x = \pm \dfrac{1}{\sqrt{3}}

Question 16: If \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b} = 2\hat{i}-\hat{j}+3\hat{k} and \vec{c}=\hat{i}-2\hat{j}+\hat{k}, find a unit vector parallel to 2\vec{a}-\vec{b}+3\vec{c} .

Solution:

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

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

=> 2\vec{a}-\vec{b}+3\vec{c} = 3\hat{i}-3\hat{j}+2\hat{k}

Thus, the unit vector is,

=> \hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}

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

=> \hat{p} = \dfrac{1}{\sqrt{22}}(3\hat{i}-3\hat{j}+2\hat{k})

Question 17: If \vec{a}=\hat{i}+\hat{j}+\hat{k} , \vec{b}=4\hat{i}-2\hat{j}+3\hat{k} and \vec{c} = \vec{i}-2\hat{j}+\hat{k}, find a vector of magnitude 6 units which is parallel to the vector 2\vec{a}-\vec{b}+3\vec{c}.

Solution:

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

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

=> 2\vec{a}-\vec{b}+3\vec{c} = \hat{i}-2\hat{j}+2\hat{k}

Unit vector in that direction is,

=> \hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}

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

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

Given that the vector has a magnitude of 6,

=> Required vectors are : \pm6\times\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k}) = \pm2(\hat{i}-2\hat{j}+2\hat{k})

Question 18: Find a vector of magnitude 5 units parallel to the resultant of the vector \vec{a}=2\hat{i}+3\hat{j}-\hat{k}and \vec{b} = \hat{i}-2\hat{j}+\hat{k}.

Solution:

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

The resultant vector will be given by,

=> \vec{r} = \vec{a}+\vec{b}

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

=> \vec{r} = 3\hat{i}+\hat{j}

Unit vector is,

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

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

=> \hat{p} = \dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})

Given that the vector has a magnitude of 5,

=> Required vectors are: \pm5\times\dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})= \pm\dfrac{5}{\sqrt{10}}(3\hat{i}+\hat{j})

Question 19: The two vectors \hat{j}+\hat{i} and 3\hat{i}+\hat{j}+4\hat{k} represent the sides \vec{AB} and \vec{AC} respectively of the triangle ABC. Find the length of the median through A.

Solution:

Let D be the point on BC, on which the median through A touches.

D is also the mid-point of BC.

The median \vec{AD} is thus given by:

=> \vec{AD} = \dfrac{\vec{B}+\vec{C}}{2}- \vec{A}

=> \vec{AD} = \dfrac{\vec{B}-\vec{A}+\vec{C}-\vec{A}}{2}

=> \vec{AD} = \dfrac{\vec{AB}+\vec{AC}}{2}

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

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

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

Thus, the length of the median is,

=> |\vec{AD}| = \sqrt{2^2+1^2+2^2}

=> |\vec{AD}| = \sqrt{9}

=> |\vec{AD}| = 3  units

Summary

Exercise 23.6 in RD Sharma's Class 12 textbook focuses on advanced vector algebra operations and identities. Key concepts covered include:

• Vector triple products and their properties
• Vector identities involving cross products
• Geometric interpretations of vector operations
• Decomposition of vectors
• Applications of dot and cross products in vector algebra

These problems require a deep understanding of vector operations and their properties, as well as the ability to manipulate and simplify complex vector expressions.