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    RD Sharma Class 12 Exercise 24.1 Vector or Cross Product Solutions Maths-Download PDF Online

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    RD Sharma Class 12 Exercise 24.1 Vector or Cross Product Solutions Maths-Download PDF Online

    Edited By Satyajeet Kumar | Updated on Jan 27, 2022 06:10 PM IST

    RD Sharma’s maths books are well known for their quality and authenticity. Still, many students find it challenging to solve the problems and lag. This problem is that a student is unable to find sources from where they can understand and learn the concepts. RD Sharma Class 12th Exercise 24.1 solutions have arrived to solve this issue with complete reliability on them.

    The chapter ‘Vector or Cross Product’ solutions are available for access at ease for students. RD Sharma Class 12th Chapter 24 Exercise 24.1 is as per the latest CBSE guidelines to help students score exceptionally in their board examinations. RD Sharma Solutions Cross Products, Unit vectors, area of parallelogram determined by the vectors, area of triangles by vectors and verify laws by vectors are discussed in this chapter. The cross-product of two vectors being the third vector perpendicular to the two original vectors is the concept discussed here. 48 questions are given in this exercise. Class 12th RD Sharma Chapter 24 Exercise 24.1 Solutions are available in PDF format and downloaded for free.

    RD Sharma Class 12 Solutions Chapter 24 Vector or Cross Product - Other Exercise

    Vector or Cross Product Excercise: 24.1

    Vector or Cross Product exercise 24.1 question 1

    Answer : \sqrt{91}
    Hint : To solve this equation by using determinate formula
    Given : \vec{a}=\hat{\imath}+3 \hat{\jmath}-2 \hat{k} \text { and } \vec{b}=\widehat{-i}+3 \hat{k}
    find \left | \vec{a}\times \vec{b} \right |
    Solution :
    \begin{gathered} \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{c} & \hat{\jmath} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{array}\right| \\ =9 \hat{i}-\hat{j}+3 \hat{k} \end{gathered}
    \begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{9^{2}+(-1)^{2}+(3)^{2}} \\\\ &=\sqrt{81+1+3} \\\\ &=\sqrt{91}\ \end{aligned}


    Vector or Cross Product exercise 24.1 question 2 (i)

    Answer : \sqrt{26}
    Hint : To solve this equation we use determinate formula then magnitude formula
    Given : \vec{a}=3 \hat{\imath}+4 \hat{\jmath}
    \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}
    Find value |\vec{a} \times \vec{b}|
    Solution :
    \begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{\jmath}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1+4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-\hat{k} \end{aligned}
    \begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{4^{2}+3^{2}+1^{2}} \\\\ &=\sqrt{16+9+1} \\\\ &=\sqrt{26} \end{aligned}


    Vector or Cross Product exercise 24.1 question 2 (ii)

    Answer :\sqrt{6}
    Hint : To solve this equation we use determinate formula then magnitude formula
    Given : \vec{a}=2 \hat{\imath}+\hat{k}
    \begin{aligned} &\vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ &\text { find }|\vec{a} \times \vec{b}| \end{aligned}
    Solution :
    \begin{aligned} : \vec{a} \times \vec{b} &=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 1-0 \times 1) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}
    \begin{aligned} |\vec{a} \times \vec{b}| &=\sqrt{1^{2}+1^{2}+2^{2}} \\\\ &=\sqrt{1+1+4} \\\\ &=\sqrt{6} \end{aligned}



    Vector or Cross Product exercise 24.1 question 3 (i)

    Answer :\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k})
    Hint : To solve this equation , use magnitude and \vec{a} \times \vec{b}
    Given :4 \hat{\imath}-\hat{\jmath}+3 \hat{k} ;-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}
    Solution :\vec{a}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}
    \vec{b}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}
    \begin{aligned} \vec{a} \times \vec{b} &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{array}\right| \\\\ &=\hat{i}(1 \times 2-3 \times 1)-\hat{j}(4 \times-2-3 \times-2)+\hat{k}(4 \times 1-(-1 \times-2)) \end{aligned}
    =-\hat{i}+2 \hat{j}+2 \hat{k}
    \begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+2^{2}+2^{2}} \\ \end{aligned}
    \begin{aligned} &=\sqrt{9} \\\\ &=3 \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}=\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}) \end{aligned}


    Vector or Cross Product exercise 24.1 question 3 (ii)

    Answer : \frac{1}{11}(\hat{\imath}+\hat{\jmath}-3 \hat{k})
    Hint : To solve this equation we use \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}
    Given : \vec{a}=2 \hat{\imath}+\hat{\jmath}+\hat{k}
    \vec{b}=\hat{\imath}+2 \hat{\jmath}+\hat{k}
    Solution : \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}
    \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|
    \begin{aligned} &=\hat{\imath}(1 \times 1-2 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 2-1 \times 1) \\\\ &=\hat{\imath}(-1)-\hat{\jmath}(1)+\hat{k}(3) \\\\ &=-\hat{\imath}-\hat{\jmath}+3 \hat{k} \end{aligned}
    |\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+3^{2}}
    \begin{aligned} &=\sqrt{11} \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{11}}(-\hat{\imath}-\hat{\jmath}+3 \hat{k}) \end{aligned}


    Vector or Cross Product exercise 24.1 question 4

    Answer : \sqrt{74}
    Hint : To solve this equation we suppose both terms in x and y then we use magnitude formula
    Given :\vec{a}=(3 \hat{k}+4 \hat{\jmath}) \times(\hat{\imath}+\hat{\jmath}-\hat{k})
    Solution :
    \begin{aligned} &\vec{x}=3 \hat{k}+4 \hat{j} \\\\ &\vec{y}=\hat{\imath}+\hat{\jmath}-\hat{k} \end{aligned}
    \begin{aligned} &|\vec{x} \times \vec{y}| \\ &\vec{x} \times \vec{y}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 0 & 4 & 3 \\ 1 & 1 & -1 \end{array}\right| \end{aligned}
    \begin{aligned} &=\hat{\imath}(4 \times-1-3 \times 1)-\hat{\jmath}(0 \times 1-3 \times 1)+\hat{k}(0 \times 1-41) \\ &=-7 \hat{\imath}+3 \hat{\jmath}-4 \hat{k} \\\\ &|\vec{x} \times \vec{y}|=\sqrt{(-7)^{2}+3^{2}+(-4)^{2}} \\\\ &=\sqrt{49+9+16} \\\\ &\vec{a}=\sqrt{74} \end{aligned}


    Vector or Cross Product exercise 24.1 question 5

    Answer : \sqrt{504}
    Hint : To solve this we multiply \vec{b}by 2 as given then magnitude formula.
    Given : \vec{a}=4 \hat{\imath}+3 \hat{\jmath}+\hat{k}
    \vec{b}=\hat{\imath}-2 \hat{k}
    Find |2 \vec{b} \times \vec{a}|
    Solution :
    We need to find unit vector of b (\hat{b})
    \begin{aligned} &\hat{b}=\frac{\vec{b}}{|\vec{b}|} \\\\ &\hat{b}=\frac{(\hat{i}-2 \hat{k})}{\sqrt{1^{2}+(-2)^{2}}} \\\\ &\hat{b}=\frac{1}{\sqrt{5}}(\hat{i}-2 \hat{k}) \end{aligned}
    \begin{array}{r} 2 \hat{b}=\frac{2}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{k}) \\\\ 2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \\\frac{2}{\sqrt{5}} & 0 & \frac{-4}{\sqrt{5}} \\\\ 4 & 3 & 1 \end{array}\right| \end{array}
    \begin{aligned} &=\hat{\mathrm{i}}\left[0.1-3 \cdot\left(\frac{-4}{\sqrt{5}}\right)\right]-\hat{\mathrm{j}}\left[\left(\frac{2}{\sqrt{5}}\right) 1-(4)\left(\frac{-4}{\sqrt{5}}\right)\right]+\mathrm{k}\left[\left(\frac{2}{\sqrt{5}}\right) 3-(4)(0)\right] \\\\ &2 \vec{b} \times \vec{a}=\frac{12}{\sqrt{5}} \hat{\mathrm{i}}-\frac{18}{\sqrt{5}} \hat{\mathrm{j}}+\frac{6}{\sqrt{5}} \hat{\mathrm{k}} \end{aligned}
    \begin{aligned} |2 \vec{b} \times \vec{a}| &=\sqrt{\left(\frac{12}{\sqrt{5}}\right)^{2}+\left(\frac{-18}{\sqrt{5}}\right)^{2}+\left(\frac{6}{\sqrt{5}}\right)^{2}} \\\\ &=\sqrt{\frac{144}{5}+\frac{324}{5}+\frac{36}{5}} \\\\ |2 \vec{b} \times \vec{a}| &=\sqrt{\frac{504}{5}} \end{aligned}



    Vector or Cross Product exercise 24.1 question 6

    Answer : -25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k}
    Hint : To solve this equation we put value in (\vec{a}+2 \vec{b}) \text { and }(2 \vec{a}-\vec{b})
    Given :\vec{a}=3 \hat{\imath}-\hat{\jmath}-2 \hat{k}
    \vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}
    Find (\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})
    Solution :
    \begin{aligned} &(\vec{a}+2 \vec{b})=(3 \hat{\imath}-\hat{\jmath}-2 \hat{k}+2(2 \hat{\imath}+3 \hat{j}+\hat{k})) \\\\ &=3 \hat{i}+4 \hat{i}-\hat{j}+6 \hat{j}-2 \hat{k}+2 \hat{k} \\\\ &=7 \hat{\imath}+5 \hat{\jmath} \end{aligned}...............(1)
    \begin{aligned} &(2 \vec{a}-\vec{b})=2(3 \hat{\imath}-\hat{\jmath}-2 \hat{k})-(2 \hat{\imath}+3 \hat{\jmath}+\hat{k}) \\\\ &=6 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}-3 \hat{\jmath}-4 \hat{k}-\hat{k} \\\\ &=4 \hat{\imath}-5 \hat{\jmath}-5 \hat{k} \end{aligned}...............(2)
    \begin{aligned} &(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{array}\right| \\\\ &=\hat{\imath}(-25-0)-\hat{\jmath}(-35-0)+\hat{k}(-35-20) \\\\ &=-25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k} \end{aligned}


    Vector or Cross Product exercise 24.1 question 7 (i)

    Answer : 42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k}
    Hint : To solve this equation we solve \vec{a} \times \vec{b} \text { then }|\vec{a} \times \vec{b}|
    Given : 2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \text { and } 3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k} magnitude=49
    Solution : \frac{\vec{a} \times \vec{b}}{(\vec{a} \times \vec{b})}=49
    \begin{aligned} &\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \\\\ &\vec{b}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{b} \\\\ &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \end{aligned}
    \begin{aligned} &=\hat{\imath}(42)-\hat{\jmath}(-14)+\hat{k}(-2) \\\\ &=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \\\\ &=7(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}
    \begin{aligned} &|\vec{a} \times \vec{b}|=7 \sqrt{6^{2}+2^{2}+(-3)^{2}} \\\\ &=7 \sqrt{36+4+9} \\\\ &=7 \sqrt{49} \quad=49 \\\\ &42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}


    Vector or Cross Product exercise 24.1 question 7 (ii)

    Answer : 2 \hat{\imath}-2 \hat{\jmath}+\hat{k}
    Hint : To solve this equation we use unit vector \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} formula
    Given :\vec{a}=3 \hat{\imath}+\hat{\jmath}-4 \hat{k} \vec{b}=6 \hat{\imath}+5 \hat{\jmath}-2 \hat{k} magnitude = 3
    Solution :3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}
    \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{array}\right|
    \begin{aligned} &=\hat{\imath}(1 \times-2-(5 \times-4))-\hat{\jmath}(3 \times-2-(-4) \times 6)+\hat{k}(3 \times 5-1 \times 6) \\\\ &=18 \hat{\imath}-18 \hat{\jmath}+9 \hat{k} \\\\ &=9(2 \hat{\imath}-2 \hat{\jmath}+\hat{k}) \end{aligned}
    \begin{aligned} &|\vec{a} \times \vec{b}|=9 \sqrt{2^{2}+(-2)^{2}+1^{2}} \\\\ &=9 \sqrt{4+4+1} \\\\ &=9 \sqrt{9} \\\\ &=9 \times 3 \end{aligned}
    \begin{aligned} &=27 \\\\ &3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{3 \times 9(2 \hat{i}-2 \hat{j}+\hat{k})}{27} \\\\ &=2 \hat{\imath}-2 \hat{j}+\hat{k} \end{aligned}



    Vector or Cross exercise 24.1 question 8 (i)

    Answer : 6 square units
    Hint : To solve this we use area of parallelogram
    Given : 2 \hat{\imath} \text { and } 3 \hat{\jmath}
    Solution : Area of parallelogram
    \begin{aligned} &=|\vec{a} \times \vec{b}| \\\\ &=|2 \hat{\imath} \times 3 \hat{\jmath}| \\\\ \end{aligned}
    \begin{aligned} &=|6 \hat{k}| \\\\ &=6 \; s q \cdot \text { units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 8 (ii)

    Answer : 3 \sqrt{3} \text { sq. units }
    Hint : To solve this , we use area of parallelogram.
    Given :2 \hat{\imath}+\hat{\jmath}+3 \hat{k} ; \hat{\imath}-\hat{\jmath}
    Solution : Area of parallelogram=|\vec{a} \times \vec{b}|
    \begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right| \\\\ &=\hat{i}(1 \times 0-3(-1))-\hat{j}(2 \times 0-3 \times 1)+\hat{k}(2 \times 1-(-1) \times 1) \\\\ &=3 \hat{\imath}+3 \hat{j}+3 \hat{k} \end{aligned}
    Area of parallelogram
    \begin{aligned} &=\sqrt{3^{2}+3^{2}+3^{2}} \\\\ &=\sqrt{9+9+9} \\\\ &=\sqrt{27} \\\\ &=3 \sqrt{3} \text { sq. units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 8 (iii)

    Answer : 10 \sqrt{3}\; s q . \text { units }
    Hint : To solve this equation we use area of parallelogram
    Given : 3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \quad \text { and } \hat{\imath}-3 \hat{\jmath}+4 \hat{k}
    Solution : Area of parallelogram =|\vec{a} \times \vec{b}|
    \begin{aligned} &\vec{a}=3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \\\\ &\vec{b}=\hat{\imath}-3 \hat{j}+4 \hat{k} \end{aligned}
    \begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{array}\right| \\\\ &=\hat{\imath}(1 \times 4-(-2)(-3))-\hat{\jmath}(3 \times 4-2 \times-1)+\hat{k}(3 \times-3-1 \times 1) \\\\ &=-2 \hat{\imath}-14 \hat{\jmath}-10 \hat{k} \end{aligned}
    \begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} \\\\ &=\sqrt{4+196+100} \\\\ &=\sqrt{300} \\\\ &=10 \sqrt{3} \text { sq.units } \end{aligned}



    Vector or Cross Product exercise 24.1 question 8 (iv)

    Answer : 4 \sqrt{2} \text { sq. units }
    Hint : To solve this equation we use area of parallelogram
    Given \hat{\imath}-3 \hat{\jmath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}
    Solution :\vec{a}=\hat{\imath}-3 \hat{\jmath}+\hat{k} \quad ; \quad \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}
    Area of parallelogram =|\vec{a} \times \vec{b}|
    \begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &\vec{a} \times \vec{b}=\hat{i}[-3.1-1.1]-\hat{j}[1.1-1.1]+\hat{k}[1.1-1 \cdot(-3)] \end{aligned}
    \begin{aligned} &=\hat{\mathrm{i}}[-3-1]-\hat{\mathrm{j}}[1-1]+\hat{\mathrm{k}}[1+3] \\\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-4 \hat{\mathbf{i}}+4 \hat{\mathrm{k}} \end{aligned}
    Area of parallelogram
    \begin{aligned} &=\sqrt{-4^{2}+4^{2}} \\\\ &=4 \sqrt{2} \end{aligned}


    Vector or Cross Product exercise 24.1 question 9 (i)

    Answer : \frac{15}{2} \text { sq.units }
    Hint : To solve this we use area of parallelogram formula
    Given : Area of parallelogram =\frac{1}{2}\left(d_{1} \times d_{2}\right)
    Solution : d_{1}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}
    \begin{aligned} &d_{2}=4 \hat{\imath}-\hat{\jmath}-3 \hat{k} \\\\ &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 1 & -2 \\ 4 & -1 & -3 \end{array}\right| \end{aligned}
    \begin{aligned} &=\hat{\imath}(1 \times-3-(-2)(-1))-\hat{j}(-2 \times-3-4 \times 2)+\hat{k}(-2 \times-1-4 \times-1) \\\\ &=-5 \mathrm{i}-14 \mathrm{j}-2 \mathrm{k} \end{aligned}
    \begin{aligned} &A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-5)^{2}+14^{2}+(-2)^{2}} \\\\ &=\frac{1}{2} \sqrt{25+196+4} \end{aligned}
    \begin{aligned} &=\frac{1}{2} \sqrt{225} \\\\ &=\frac{15}{2} \text { sq.units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 9 (ii)

    Answer : \frac{\sqrt{6}}{2} \text { sq.units }
    Hint : To solve this we use area of parallelogram formula
    Given : 2 \hat{\imath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}
    Solution : Area of parallelogram =\frac{1}{2}\left(d_{1} \times d_{2}\right)
    \begin{gathered} d_{1}=2 \hat{\imath}+\hat{k} \\\\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}
    \begin{aligned} &d_{1} \times d_{2}\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}
    \begin{aligned} &\frac{1}{2}\left|d_{1} \times d_{2}\right|=\frac{1}{2} \sqrt{(-1)^{2}+(-1)^{2}+2^{2}} \\\\ &=\frac{1}{2} \sqrt{1+1+4} \\\\ &=\frac{\sqrt{6}}{2} \text { sq.units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 9 (iii)

    Answer : \frac{\sqrt{26}}{2} \text { sq.units }
    Hint : To solve this we use area of parallelogram formula
    Given : 3 \hat{\imath}+4 \hat{\jmath} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}
    Solution :Area of parallelogram =\frac{1}{2}\left(d_{1} \times d_{2}\right)
    \begin{gathered} d_{1}=3 \hat{\imath}+4 \hat{\jmath} \\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}
    \begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{j}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1-4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-1 \hat{k} \end{aligned}
    \begin{aligned} &A \quad=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-4)^{2}+(-3)^{2}+(-1)^{2}} \\\\ &=\frac{1}{2} \sqrt{16+9+1} \\\\ &=\frac{\sqrt{26}}{2} \text { sq.units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 9 (iv)

    Answer : \frac{49}{2} \text { sq.units }
    Hint : To solve this we use area of parallelogram formula
    Given : d_{1}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \quad, d_{2}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}
    Solution : Area of parallelogram =\frac{1}{2}\left(d_{1} \times d_{2}\right)
    \begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \\\\ \quad &=\hat{\imath}(3 \times 2-(-6 \times 6))-\hat{\jmath}(2 \times 2-3 \times 6)+\hat{k}(2 \times-6-3 \times 3) \\\\ &\quad=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}
    \begin{gathered} A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ =\frac{1}{2} \sqrt{42^{2}+14^{2}+-21^{2}} \\\\ =\frac{1}{2} \sqrt{144+196+9} \end{gathered}
    \begin{aligned} &=\frac{1}{2} \sqrt{349} \\\\ &=\frac{49}{2} \text { sq.units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 10

    Answer : not equal
    Hint : To solve this w use determinant method
    Given :
    \begin{aligned} &\vec{a}=2 \hat{\imath}+5 \hat{\jmath}-7 \hat{k} \\ &\vec{b}=3 \hat{i}+4 \hat{\jmath}+\hat{k} \\ &\vec{c}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k} \end{aligned}
    Solution : \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{j} & \hat{k} \\ 2 & 5 & -7 \\ -3 & 4 & 1 \end{array}\right|
    \begin{aligned} &\vec{a}=33 \hat{\imath}+19 \hat{\jmath}+23 \hat{k} \\\\ &\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -3 & 4 & 1 \\ 1 & -2 & -3 \end{array}\right| \\\\ &\vec{l}=-10 \hat{\imath}-8 \hat{\jmath}+2 \hat{k} \end{aligned}
    \begin{aligned} &\vec{d} \times \vec{c}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 33 & 19 & 23 \\ 1 & -2 & -3 \end{array}\right| \\\\ &(\vec{a} \times \vec{b}) \times \vec{c}=-11 \hat{\imath}+122 \hat{\jmath}-85 \hat{k} \\\\ &\vec{a} \times \vec{l}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 5 & -7 \\ -10 & -8 & 2 \end{array}\right| \end{aligned}
    \begin{aligned} &\vec{a} \times(\vec{b} \times \vec{c})=-46 \hat{\imath}+66 \hat{\jmath}+34 \hat{k} \\\\ &(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a}(\vec{b} \times \vec{c}) \end{aligned}


    Vector or Cross Product exercise 24.1 question 11

    Answer : 6

    Hint : To solve this we use \vec{a}\; \&\; \vec{b} formula
    Given : |\vec{a}|=2 ;|\vec{b}|=5 \text { and }(\vec{a} \times \vec{b})=8 \text { find } \vec{a} \cdot \vec{b}
    Solution : |\vec{a} \times \vec{b}|=|\vec{a}| \cdot|\vec{b}| \sin \theta
    \begin{aligned} &8=2 \times 5 \sin \theta \\\\ &8=10 \sin \theta \\\\ &\sin \theta=\frac{8}{10}=\frac{4}{5} \end{aligned}
    \begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &=2 \times 5 \times \frac{3}{5} \\\\ &=6 \end{aligned}


    Vector or Cross Product exercise 24.1 question 12

    Answer : These are unit vectors as well as perpendicular
    Hint : To solve this , we do magnitude of one by one
    Given :
    \begin{aligned} &\vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\\\ &\vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\\\ &\vec{c}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}

    Solution : |\vec{a}|=\frac{1}{7} \sqrt{2^{2}+3^{2}+6^{2}}

    \begin{aligned} &=\frac{1}{7} \sqrt{49}=>\frac{7}{7}=1 \\\\ &|\vec{b}|=\frac{1}{7} \sqrt{49}=1 \\\\ &|\vec{c}|=\frac{1}{7} \sqrt{36+4+9} \end{aligned}
    \begin{aligned} &=\frac{1}{7} \sqrt{49}=\frac{7}{7}=1 \\\\ &\vec{a} \cdot \vec{b}=0 \\\\ &\vec{b} \cdot \vec{c}=0 \end{aligned}
    \begin{aligned} &\vec{a} \cdot \vec{b}=\frac{1}{49}[6-18+12] \\\\ &\vec{b} \cdot \vec{c}=\frac{1}{49}[18-12-6] \end{aligned}


    Vector or Cross Product exercise 24.1 question 13

    Answer : 25
    Hint : To solve this formula \vec{a} \cdot \vec{b} \text { and }|\vec{a} \times \vec{b}|
    Given : |\vec{a}|=13 \quad ;|\vec{b}|=5
    \vec{a} \cdot \vec{b}=60
    Solution : \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta
    \begin{aligned} &60=(13)(5) \cos \theta \\\\ &\cos \theta=\frac{60}{13 \times 5}=\frac{12}{13} \end{aligned}
    \begin{aligned} &\cos ^{2} \theta+\sin ^{2} \theta=1 \\\\ &\sin \theta=\sqrt{1+\cos ^{2} \theta} \\\\ &=\sqrt{1-\left(\frac{12}{13}\right)^{2}} \end{aligned}
    \begin{aligned} &=\sqrt{1-\frac{144}{169}} \\\\ &=\sqrt{\frac{169-144}{169}} \\\\ &=\sqrt{\frac{25}{169}} \end{aligned}
    \begin{aligned} &=\frac{5}{13} \\\\ &|\vec{a} \times \vec{b}|=|| \vec{a}|| \vec{b}|\sin \theta| \\\\ &=13 \times 5 \times \frac{5}{13} \\\\ &=25 \end{aligned}


    Vector or Cross Product exercise 24.1 question 14

    Answer :\frac{\pi }{4}
    Hint : To solve this equation we use aband |\vec{a}||\vec{b}| \text { and }|\vec{a} \times \vec{b}|
    Given : |\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}
    Solution :
    1)\vec{a} \times \vec{b}|=|a||b| \cos \theta\\
    2) \vec{a} \cdot \vec{b}|=|a||b| \sin \theta
    \Rightarrow|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta
    \begin{aligned} &\Rightarrow \frac{\sin \theta}{\cos \theta}=1 \\\\ &\Rightarrow \tan \theta=1 \\\\ &\Rightarrow \theta=\tan ^{-1} 1 \\\\ &\theta=\frac{\pi}{4} \end{aligned}


    Vector or Cross Product exercise 24.1 question 15

    Answer : \vec{a}+\vec{c}=m \vec{b}
    Hint : Here m is the constant team
    Given :\vec{a} \times \vec{b}=\vec{b} \times \vec{c}
    \vec{a} \times \vec{b}-\vec{b} \times \vec{c}=0
    Solution :
    \begin{aligned} &\vec{b} \times \vec{c}=-\vec{c} \times \vec{b} \text { (substitute) } \\\\ &\vec{a} \times \vec{b}+\vec{c} \times \vec{b}=0 \\\\ &(\vec{a}+\vec{c}) \times \vec{b}=0 \\\\ &\vec{a}+\vec{c}=m \vec{b} \end{aligned}


    Vector or Cross Product exercise 24.1 question 16

    Answer : \frac{\pi }{6}
    Hint : To solve this we use |\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta formula
    Given : \vec{a} \times \vec{b}=3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k}
    Solution : |\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta
    \begin{gathered} \sqrt{3^{2}+2^{2}+6^{2}}=2 \times 7 \sin \theta \\\\ \sqrt{9+4+12}=14 \sin \theta \\\\ 7 =14 \sin \theta \end{gathered}
    \begin{aligned} \sin \theta &=\frac{1}{2} \\\\ \theta &=\frac{\pi}{6} \end{aligned}


    Vector or Cross Product exercise 24.1 question 17

    Answer : \vec{a} \cdot \vec{b}=0 ; \vec{a} \times \vec{b}=0
    Hint : To solution we know \vec{a}, \vec{b} are parallel
    Given :\vec{a} \times \vec{b}=0
    Solution : 1) \vec{a} and \vec{b} are parallel
    Or
    \begin{aligned} &|\vec{a}|=0 \text { or }|\vec{b}|=0 \\\\ &\vec{a} \cdot \vec{b}=0 \end{aligned}
    1. \vec{a} & \vec{b} are perpendicular
    2. |\vec{a}|=0 \text { or }|\vec{b}|=0
    Either |\vec{a}|=0 \text { or }|\vec{b}|=0 or both are zero if \vec{a} \cdot \vec{b}=0\; \& \; \vec{a} \times \vec{b}=0




    Vector or Cross Product exercise 24.1 question 18

    Answer:

    Hint : To solve this equation we use |\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta
    Given : \vec{a} \times \vec{b}=\vec{c} \text { and } \vec{b} * \vec{c}=\vec{a} \text { and } \vec{c} * \vec{a}=\vec{b}
    Solution : \vec{a} \times \vec{b}=\vec{c}
    \begin{aligned} &\Rightarrow \vec{a} \times \vec{b}=\vec{c} \\\\ &\Rightarrow|\vec{a}||\vec{b}| \sin \theta=\vec{c} \\\\ &\theta=\sin ^{-1}(1) \\\\ &\theta=\frac{\Pi}{2} \; 0 r \; 90^{\circ} \end{aligned}

    Similarly can be prove for others

    \vec{b} \times \vec{c}=\vec{a} this means perpendicular (\vec{b} \times \vec{c})
    \vec{c} \times \vec{a}=\vec{b} this means perpendicular (\vec{c} \times \vec{a})
    This are together \vec{a}, \vec{b} \text { and } \vec{c} form orthonormal trial


    Vector or Cross Product exercise 24.1 question 19

    Answer : \frac{1}{\sqrt{165}}(10 \hat{\imath}+7 \hat{\jmath}-6 \hat{k})
    Hint : To solve this equation , we use determination method
    Given : A=(3,-1,2)

    \begin{aligned} &B=(1,-1,-3) \\\\ &C=(4,-3,1) \end{aligned}
    Solution : \overrightarrow{A B}=-2 \hat{\imath}-5 \hat{k}
    \begin{aligned} &\overrightarrow{A C}=\hat{\imath}-2 \hat{\jmath}-\hat{k} \\\\ &\vec{C}=x \overrightarrow{A B}+y \overrightarrow{A C} \end{aligned}
    If \vec{a} is perpendicular to \vec{c}
    \vec{d} is perpendicular to \overrightarrow{AB} ,\vec{d} perpendicular to \overrightarrow{AC}
    \begin{aligned} &\vec{d}=\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{array}\right| \\\\ &=-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k} \\\\ &\hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k}}{\sqrt{165}} \end{aligned}


    Vector or Cross Product exercise 24.1 question 20

    Answer:
    To have \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
    Hint:
    To solve this we use |\vec{a} \times \vec{b}| \text { and }|\vec{a}||\vec{b}| formula
    Given:
    Solution : \overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B}=0
    \begin{aligned} &\overrightarrow{A B} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{A B} \times 0 \\\\ &a c(\sin B-\sin A)=0 \end{aligned}
    Divide by abc
    \begin{aligned} &\frac{\sin B}{b}-\frac{\sin A}{a}=0 \\\\ &\frac{\sin B}{b}=\frac{\sin A}{a} \end{aligned} .......(i)
    \begin{aligned} &\overrightarrow{B C} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin B}{b}=\frac{\sin C}{c} \end{aligned}.......(ii)
    \begin{aligned} &\overrightarrow{C A} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin C}{c}=\frac{\sin A}{a} \end{aligned}.......(iii)
    From (i),(ii) and (iii)

    \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}


    Vector or Cross Product exercise 24.1 question 21

    Answer: \hat{i}+11 \hat{j}+7 \hat{k}
    Hint:
    To solve this we use determinant method
    Given:
    \begin{aligned} &\vec{c}=\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{array}\right| \\\\ &=\hat{i}+11 \hat{j}+7 \hat{k} \end{aligned}
    Solution:
    \begin{aligned} &\hat{i}+11 \hat{j}+7 \hat{k} \\\\ &\vec{c} \cdot \vec{a}=1-22+21=0 \\\\ &\vec{c} \text { is } \perp \text { to } \vec{a} \end{aligned}


    Vector or Cross Product exercise 24.1 question 22

    Answer: \frac{3}{4} Square units
    Hint: use concept.
    Given:
    A=\frac{1}{2}|\vec{a} \times \vec{b}|
    Solution:
    \begin{aligned} &A=\frac{1}{2}|\vec{p}+2 \vec{q}| \times|2 \vec{p}+\vec{q}| \\\\ &=\frac{1}{2}|0| \times \vec{p} \times \vec{q}+4|\vec{p}+\vec{q}|+0 \\\\ &\{\vec{p} \times \vec{p}=0\} \end{aligned}
    \begin{aligned} &=\frac{1}{2}|\vec{q} \times \vec{p}+4 \vec{q} \times \vec{p}| \\\\ &=\frac{1}{2} \times 3|\vec{q} \times \vec{p}| \end{aligned}
    \begin{aligned} &=\frac{3}{2}|\vec{p} \| \vec{q}| \sin \theta \\\\ &=\frac{3}{2} \cdot \frac{1}{2} \end{aligned}
    =\frac{3}{4} Square units


    Vector or Cross Product exercise 24.1 question 23

    Answer:
    L.H.S = R.H.S
    Hint:
    To solve this we use formula
    Given:
    |\vec{a} \times \vec{b}|^{2}=\left|\begin{array}{ll} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right|
    Solution:
    RHS \\\\\begin{aligned} &\left|\begin{array}{ll} \left|\vec{a}^{2}\right| & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right| \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) \end{aligned}
    =|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}]
    \begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2}-(|\vec{a}||\vec{b}| \cos \theta)^{2} \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \end{aligned}
    \begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &=|\vec{a} \times \vec{b}|^{2} \quad[\because \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta] \end{aligned}


    Vector or Cross Product exercise 24.1 question 24

    Answer: |\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta
    Hint:
    Given: |\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta
    Solution:
    |\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta .........(1)
    Also,
    |\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}| \cos \theta ............(2)
    Dividing equation (1) and (2)
    \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=\frac{|\vec{a}| \vec{b} \mid \sin \theta}{|\vec{a}||\vec{b}| \cos \theta}
    \begin{aligned} &\frac{|\vec{a} \times \vec{b}|}{(\vec{a} \cdot \vec{b})}=\tan \theta \\\\ &|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta \end{aligned}


    Vector or Cross Product exercise 24.1 question 25

    Answer: 7
    Hint:
    To solve this we use |\vec{a} \times \vec{b}| formula
    Given:
    \begin{aligned} &|\vec{a}|=\sqrt{26} ;|\vec{b}|=7 \\\\ &(\vec{a} \times \vec{b})=35 \end{aligned}
    Solution:
    \begin{aligned} &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \\\\ &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \end{aligned}
    \begin{aligned} &(35)^{2}=(\sqrt{26})^{2}(7)^{2}-(\vec{a} \cdot \vec{b})^{2} \\\\ &|\vec{a} \cdot \vec{b}|=\sqrt{49} \\\\ &\vec{a} \cdot \vec{b}=7\ \end{aligned}

    Vector or Cross Product exercise 24.1 question 26

    Answer:
    3 \sqrt{5}square units
    Hint:
    To solve this we use area of triangle formula
    Given:
    \begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ &\overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \end{aligned}
    Solution:
    \begin{aligned} &A=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{O B}| \\\\ &=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{array}\right| \end{aligned}
    \begin{aligned} &=\frac{1}{2}|8 \hat{i}-10 \hat{j}+4 \hat{k}| \\\\ &=\frac{1}{2} \sqrt{64+100+16} \end{aligned}
    =\frac{\sqrt{180}}{2}=\sqrt{45}\\\\
    =3 \sqrt{5} square units


    Vector or Cross Product exercise 24.1 question 27 (i)

    Answer:
    \frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k})
    Hint: To solve this we use determinant method
    Given:
    \begin{aligned} &\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} \\\\ &\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} \\\\ &\vec{c} \cdot \vec{d}=15 \end{aligned}
    Solution:
    D is perpendicular to a and b both. Hence, parallel to a*b
    \begin{aligned} &\vec{a} \times \vec{b}=\vec{c} \\\\ &\vec{a} \times \vec{b}=\vec{d} \\\\ &\vec{d}=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{array}\right| \end{aligned}
    \begin{aligned} &\vec{d}=\lambda(32 \hat{i}-\hat{j}-14 \hat{k}) \\\\ &\vec{c} \cdot \vec{d}=15 \\\\ &\lambda(2 \hat{i}-\hat{j}+4 \hat{k})(32 \hat{i}-\hat{j}-14 \hat{k})=15 \end{aligned}
    \begin{aligned} &\lambda(64+1-56)=15 \\\\ &9 \lambda=15 \\\\ &\lambda=\frac{5}{3} \\\\ &\vec{d}=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \end{aligned}


    Vector or Cross Product exercise 24.1 question 27 (ii)

    Answer:
    \vec{a}=7(\hat{i}-\hat{j}-\hat{k})
    Hint:
    To solve this question we suppose term in terms of x,y,z.
    Given:
    \begin{aligned} &\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k} \\\\ &\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k} \\\\ &\vec{c}=3 \hat{i}+\hat{j}-\hat{k} \\\\ &\vec{d} \cdot \vec{c}=21 \end{aligned}
    Solution: Let \vec{d}=x \hat{i}+y \hat{j}+z \hat{k}
    Now, \vec{d} \cdot \vec{c}=0
    \begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})=0 \\\\ &3 x+y-z=0 \end{aligned}.......(i)
    Now, \vec{d} \cdot \vec{b}=0
    \begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})=0 \\\\ &x-4 y+5 z=0 \end{aligned}....(ii)
    Now,
    \begin{aligned} &\vec{d} \cdot \vec{a}=0 \\\\ &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})=0 \\\\ &4 x+5 y-z=0 \end{aligned}....(iii)
    Solving (2) and (3)
    \begin{aligned} &\frac{x}{25-4}=\frac{y}{-1-20}=\frac{z}{-16-5}=k \\\\ &\frac{x}{-21}=\frac{y}{-21}=\frac{z}{-21}=k \end{aligned}
    \begin{aligned} &\frac{x}{1}=\frac{y}{-1}=\frac{z}{-1}=k \\\\ &x=1 ; y=-1 ; z=-k \end{aligned}
    Putting value in (i)
    \begin{aligned} &3 k-k+k=21 \\\\ &k=7 \\\\ &x=7 ; y=-7 ; z=-7 \\\\ &\vec{a}=7(\hat{i}-\hat{j}-\hat{k}) \end{aligned}

    Vector or Cross Product exercise 24.1 question 28

    Answer:
    \frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k})
    Hint:
    To solve this we use determinant method
    Given:
    \begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}
    Solution:
    \begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}
    Let
    \begin{aligned} &\vec{d}=\vec{a}+\vec{b} \\\\ &\vec{d}=4 \hat{i}+4 \hat{j}+0 \hat{k} \end{aligned}
    And
    \begin{aligned} &\vec{e}=\vec{a}-\vec{b} \\\\ &=2 \hat{i}+0 \hat{j}+4 \hat{k} \end{aligned}
    Let \vec{f} be any vector perpendicular to both \vec{d} \; \& \; \vec{e}, hence parallel to \vec{d} \times \vec{e}
    \begin{aligned} &\therefore \vec{f}=\vec{d} \times \vec{e} \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array}\right| \\\\ &=16 \hat{i}-16 \hat{j}-8 \hat{k} \end{aligned}
    \begin{aligned} &=\frac{16\hat {i}-16 \hat{j}-8 \hat {k}}{\sqrt{(16)^{2}+(-16)^{2}+(-8)^{2}}} \\\\ &=\frac{16}{24} \hat{i}-\frac{16}{24} \hat{j}-\frac{8}{24} \hat{k} \\\\ &=\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k}) \end{aligned}


    Vector or Cross Product exercise 24.1 question 29

    Answer:
    \frac{\sqrt{61}}{2} \text { square units }
    Hint:
    To solve this we use area of triangle ABC
    Given:
    \begin{aligned} &A(2,3,5) \\ &B(3,5,8) \\ &C(2,7,8) \end{aligned}
    \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|
    Solution:
    \begin{aligned} &\overrightarrow{A B}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ &\overrightarrow{B C}=-\hat{i}+2 \hat{j} \\\\ &\overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}
    \begin{aligned} &=\hat{i}(0-6)-\hat{j}(0+3)+\hat{k}(2+2) \\\\ &=-6 \hat{i}-3 \hat{j}+4 \hat{k} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \end{aligned}
    \begin{aligned} &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{36+9+16} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{61}\\\\ \end{aligned}
    Area \; of \; \Delta \mathrm{ABC}=\frac{1}{2}|\sqrt{61}|
    =\frac{\sqrt{61}}{2} square\; units


    Vector or Cross Product exercise 24.1 question 30

    Answer:
    \frac{\sqrt{21}}{2} square units
    Hint:
    To solve this we use area of triangle
    Given:
    \begin{aligned} &\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k} \\\\ &\vec{b}=-\hat{i}+\hat{k} \\\\ &\vec{c}=2 \hat{j}-\hat{k} \end{aligned}
    \begin{aligned} &\vec{a}+\vec{b}=\hat{i}-3 \hat{j}+2 \hat{k} \\\\ &\vec{b}+\vec{c}=-\hat{i}+2 \hat{j} \end{aligned}
    Solution:
    \begin{aligned} &A=\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\\\ &A=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}
    \begin{aligned} &=\frac{1}{2}|-4 \hat{i}-2 \hat{j}-\hat{k}| \\\\ &=\frac{1}{2} \sqrt{16+4+1} \\\\ &=\frac{\sqrt{21}}{2} \text { square units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 31

    Answer:
    11 \sqrt{5} \text { square units }
    Hint:
    To solve this we use \vec{c}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}
    Given:
    \begin{aligned} &\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \\\\ &\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k} \end{aligned}
    Solution:
    \begin{aligned} &\vec{c}=\vec{a}+\vec{b}=3 \hat{i}-6 \hat{j}+2 \hat{k} \\\\ &\vec{c}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \\\\ &=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k} \end{aligned}
    \begin{aligned} &A=|\vec{a} \times \vec{b}|=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{array}\right| \\\\ &=|22 \hat{i}+11 \hat{j}+0 \hat{k}| \\\\ &=11|2 \hat{i}+\hat{j}| \end{aligned}
    \begin{aligned} &=11 \sqrt{(2)^{2}+(1)^{2}} \\\\ &=11 \sqrt{5} \end{aligned}


    Vector or Cross Product exercise 24.1 question 32

    Answer:
    No, take any two collinear vectors
    Hint:
    To solve this we let
    \hat{i}+\hat{j}+\hat{k}=\overrightarrow{0} ; 2 \hat{i}+2 \hat{j}+2 \hat{k}=\overrightarrow{0}
    Given: \vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0
    Solution:
    Statement:
    If either \vec{a}=\vec{b} \text { or } \vec{b}=\overrightarrow{0} \Rightarrow \vec{a} \times \vec{b}=0
    Converse: \vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0
    Let
    \begin{aligned} &\vec{a}=\hat{i}+\hat{j}+\hat{k}=(1,1,1) \\\\ &\vec{b}=2 \hat{i}+2 \hat{j}+2 \hat{k}=(2,2,2) \end{aligned}
    \begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right| \\\\ &=\hat{i}(2-2)+\hat{j}(2-2)+\hat{k}(2-2) \\\\ &=\vec{o} \end{aligned}
    But \vec{a} \neq 0 \text { and } \vec{b} \neq 0


    Vector or Cross Product exercise 24.1 question 33

    Answer:
    Proved
    Hint:
    To solve this we use determinant method
    Given:
    \begin{aligned} &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} \end{aligned}
    Solution:
    \vec{b}+\vec{c}=\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+\left(b_{3}+c_{3}\right) \hat{k}
    \vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{array}\right|
    =\hat{i}\left[a_{2}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{2}+c_{2}\right)\right]-\hat{j}\left[a_{1}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{1}+c_{1}\right)\right]+\hat{k}\left[a_{1}\left(b_{2}+c_{2}\right)-a_{2}\left(b_{1}+c_{1}\right)\right]
    \begin{aligned} &=\hat{i}\left(a_{2} b_{3}+a_{3} b_{2}\right)-\hat{j}\left(a_{1} b_{3}+a_{3} b_{1}\right)+\hat{k}\left(a_{1} b_{2}+a_{2} b_{1}\right)+\hat{i}\left(a_{2} c_{3}+a_{3} c_{2}\right)-\hat{j}\left(a_{1} c_{3}+a_{3} c_{1}\right)+\hat{k}\left(a_{1} c_{2}+a_{2} c_{1}\right) \\\\ &\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{c})+(\vec{a} \times \vec{b}) \end{aligned}



    Vector or Cross Product exercise 24.1 question 34

    Answer:
    \sqrt{61} Square units
    Hint:
    To solve this we use area of triangle
    Given:
    A(1,1,2) ; B(2,3,5) ; C(1,5,5)
    Solution:
    \begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}+3 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k}) \\\\ &=\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}
    \begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+5 \hat{j}+5 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k}) \\\\ &=-\hat{i}+2 \hat{j} \end{aligned}
    \begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}
    \begin{aligned} &=\hat{i}(0-6)+\hat{j}(0-(-3))+\hat{k}(2-(-2)) \\\\ &=-6 \hat{i}-3 \hat{j}+4 \hat{k} \end{aligned}
    \begin{aligned} &=(\overrightarrow{A B} \times \overrightarrow{B C})=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \\\\ &=\sqrt{36+9+6} \\\\ &=\sqrt{61} \text { square units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 35

    Answer:
    \frac{1}{2} \sqrt{274} Square units
    Hint:
    To solve this we use area of triangle
    Given:
    A(1,2,3) ; B(2,-1,4) ; C(4,5,-1)
    Solution: D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})
    \begin{aligned} &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\\\ &=\hat{i}-3 \hat{j}+\hat{k} \end{aligned}
    \begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k}) \\\\ &=3 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned}
    \begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \end{aligned} \mid
    =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right|
    \begin{aligned} &=\hat{i}(12-3)+\hat{j}(-4-3)+\hat{k}(3+9) \\\\ &=9 \hat{i}+7 \hat{j}+12 \hat{k} \\\\ &=(\overrightarrow{A B} \times \overrightarrow{A C})=\sqrt{81+49+144} \\\\ &=\frac{1}{2} \sqrt{274} \text { square units } \end{aligned}

    Vector or Cross Product exercise 24.1 question 36

    Answer:
    \frac{2 \hat{i}-\hat{j}-\hat{k}}{\sqrt{6}},-\frac{3}{5} \hat{j}-\frac{4}{5} \hat{k}, \sqrt{404} \text { sq.units }
    Hint:
    To solve this we use determinant method
    Given:
    \begin{aligned} &\text { Magnitude }=10 \sqrt{3} \\ &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}
    Solution:
    \begin{aligned} &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\\\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \\\\ &\vec{a}+\vec{b}=\vec{c}=4 \hat{i}-2 \hat{j}-2 \hat{k} \end{aligned}
    \begin{aligned} &|\vec{a} \times \vec{b}|=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{(4)^{2}+(-2)^{2}+(-2)^{2}}} \\\\ &=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{16+4+4}}=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{24}} \end{aligned}
    \begin{aligned} &=\frac{4 \hat{i}-2 \hat{i}-2 \hat{k}}{2 \sqrt{6}} \\\\ &=\frac{2 \hat{i}-j-\hat{k}}{\sqrt{6}} \end{aligned}
    Similarly
    \begin{aligned} &\vec{d}=\vec{a}-\vec{b}=0 \hat{i}-6 \hat{j}-8 \hat{k} \\\\ &|\vec{a}-\vec{b}|=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{(-6)^{2}+(-8)^{2}}} \\\\ &=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{36+64}} \end{aligned}
    \begin{aligned} &=\frac{-6 \hat{j}-8 \hat{k}}{10} \\\\ &=\frac{-3}{5} \hat{j}-\frac{4}{5} \hat{k} \end{aligned}
    Area of the parallelogram =|\vec{a} \times \vec{b}|
    =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & -5 \\ 2 & 2 & 3 \end{array}\right|
    \begin{aligned} &=|-2 \hat{i}-16 \hat{j}+12 \hat{k}| \\\\ &=\sqrt{(-2)^{2}+(-16)^{2}+(12)^{2}} \\\\ &=\sqrt{4+256+144} \\\\ &=\sqrt{404} \text { sq.units } \end{aligned}


    Vector or Cross Product exercise 24.1 question 37

    Answer:
    4
    Hint: To solve this we use |\vec{a} \times \vec{b}|,|\vec{a} \cdot \vec{b}| formula
    Given: |\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=400
    Solution:
    \begin{aligned} &|\vec{a} \times \vec{b}|=|\vec{a}| \times|\vec{b}| \sin \theta \\\\ &|\vec{a} \cdot \vec{b}|=|\vec{a}| \times|\vec{b}| \cos \theta \\\\ &|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=400 \end{aligned}
    \begin{aligned} &|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=400 \\\\ &|\vec{a}|^{2}|\vec{b}|^{2}=400 \\\\ &|\vec{b}|^{2}=\frac{400}{|\vec{a}|^{2}}=\frac{400}{5^{2}}=\frac{400}{25} \end{aligned}
    \begin{aligned} &|\vec{b}|^{2}=16 \\\\ &|\vec{b}|=4 \end{aligned}


    Vector or Cross Product exercise 24.1 question 38

    Answer: \frac{\sqrt{24}}{7}
    Hint: To solve this we use \sin ^{2} \theta+\cos ^{2} \theta=1 formula
    Given:
    \begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k} \end{aligned}
    Solution:
    \begin{aligned} &\cos \theta=\vec{u} \cdot \vec{v}=|\vec{u}||\vec{v}| \cos \theta \\\\ &\cos \theta=\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \end{aligned}
    \begin{aligned} &\cos \theta=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot(3 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^{2}+(-2)^{2}+(3)^{2}} \sqrt{3^{2}+(-2)^{2}+(1)^{2}}} \\\\ &\cos \theta=\frac{3+4+3}{\sqrt{14} \sqrt{14}}=\frac{10}{14} \end{aligned}
    \begin{aligned} &\cos \theta=\frac{5}{7} \\\\ &\sin ^{2} \theta+\cos ^{2} \theta=1 \\\\ &\sin ^{2} \theta=1-\cos ^{2} \theta \\\\ &\sin ^{2} \theta=1-\left(\frac{5}{7}\right)^{2} \end{aligned}
    \begin{aligned} &\sin ^{2} \theta=\frac{49-25}{49} \\\\ &\sin ^{2} \theta=\frac{24}{49} \\\\ &\sin \theta=\sqrt{\frac{24}{49}}=\frac{\sqrt{24}}{7} \end{aligned}


    Experts prepare Class 12th RD Sharma Chapter 24 Exercise 24.1 Solutions using the latest CBSE guidelines and marking schemes. When a student refers to RD Sharma Class 12th Solutions Vector or Cross Product Ex. 24.1, it helps them have a clear understanding of the topic. When a student relates to RD Sharma Class 12th Exercise 24.1, the following benefits will accrue:

    1. Expertly crafted:

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    A team of subject experts creates RD Sharma Class 12 Solutions Chapter 24 ex 24.1 in accordance with the most recent marking pattern and CBSE guidelines. Every student must practise RD Sharma Class 12th Exercise 24.1 in order to achieve a high score. RD Sharma Class 12th Chapter 24 Exercise 24.1 solutions explain every topic so that every student can learn and solve it.

    2. The best source of preparation:

    Because of their high quality, RD Sharma Class 12th Exercise 24.1 solutions are the best to use. The most recent NCERT and CBSE updates make it an excellent choice. Furthermore,if the practice is done regularly, it would be easier to score good marks.

    3. Questions from the NCERT:

    The NCERT is usually used to set questions in exams. Teachers assign question papers based on NCERT textbooks, so it is necessary to learn NCERT concepts and solve questions from RD Sharma Class 12 Solutions.

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    When a team develops a solution,multiple approaches towards a single question develops making it easier for a student to understand the concept. The experts have devised novel approaches to answering a single question, making us unique.

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    Students will find all of these solutions for free at Career360, the best website to visit for all of the benefits associated with board exam questions and competitive exams. With Career360, a student will understand, learn, and perform exceptionally well in exams. These solutions have benefited thousands of students, making Career360 the ideal choice for all students. If you still have a doubt, visit the website and access it once. You will surely love it.

    RD Sharma Chapter-wise Solutions

    Frequently Asked Question (FAQs)

    1. What is the formula cross product?

    If two vectors A and B have an angle θ, then the formula for the cross product of vectors is given by:

    A × B = |A| |B| sin θ

    2. What do you mean by a cross product?

    A Cross product is a binary operation on two vectors in three-dimensional space.

    3. How can I avail the benefit from these solutions?

    You can visit the website, download the PDF for free and practice regularly to score good marks and avail full benefits.

    4. Will the RD Sharma Solutions for Class 12 help to score brilliant marks in the board exams?

    Yes, the RD Sharma Solutions for Class 12 is one of the best study materials which are available online. Chapter-wise and Exercise-wise solutions can be referred to by the students when they are not able to find or understand an accurate answer for the textbook questions.

    5. How can I get an understanding of the key concepts in the RD Sharma Class 12th Exercise 24.1?

    Before the start of an academic year, every student should download and read the CBSE syllabus to get a good understanding of the concepts that will be covered in the exams. 

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    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

    Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Producer

    An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

    They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

    2 Jobs Available
    Fashion Blogger

    Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

    Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Content Writer

    Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

    2 Jobs Available
    Reporter

    Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

    2 Jobs Available
    Linguist

    Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

    Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

    2 Jobs Available
    Welding Engineer

    Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

    5 Jobs Available
    QA Manager
    4 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager
    3 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

    2 Jobs Available
    Merchandiser
    2 Jobs Available
    QA Manager
    4 Jobs Available
    Azure Administrator

    An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

    4 Jobs Available
    AWS Solution Architect

    An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

    4 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager
    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Big Data Analytics Engineer

    Big Data Analytics Engineer Job Description: A Big Data Analytics Engineer is responsible for collecting data from various sources. He or she has to sort the organised and chaotic data to find out patterns. The role of Big Data Engineer involves converting messy information into useful data that is clean, accurate and actionable. 

    2 Jobs Available
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