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Vector Algebra Test - 43

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Vector Algebra Test - 43
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  • Question 1
    1 / -0
    If 2ab=a b  2\overrightarrow { a }\cdot \overrightarrow { b } =\left| \overrightarrow { a }  \right| \left| \overrightarrow { b }  \right|  then the angle between a\vec { a } and  b \vec { b } is 
    Solution
    2ab=a b  2\overrightarrow { a }\cdot \overrightarrow { b } = \left| \overrightarrow { a }  \right| \left| \overrightarrow { b }  \right| 
    But we know that ab=abcosθ\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos\theta
    cosθ=12\therefore \cos\theta=\dfrac{1}{2}
    θ=60o\Rightarrow \theta=60^o

    Here θ\theta is the angle between both the vectors 
  • Question 2
    1 / -0
    For non-zero vectors a\overrightarrow{a} and b\overrightarrow{b} if a+b<ab|\overrightarrow{a}+\overrightarrow{b}| < |\overrightarrow{a}-\overrightarrow{b}|, then a\overrightarrow{a} and b\overrightarrow{b} are
    Solution
    a+b<ab|\vec{a}+\vec{b}|<|\vec{a}-\vec{b}|
    a2+b2+2abcosθ<a2+b22abcosθa^{2}+b^{2}+2ab\cos\theta<a^{2}+b^{2}-2ab\cos\theta       where θ\theta is the included angle between the two vectors.
    Hence
    4abcosθ<04ab\cos\theta<0
    Since a,b>0a,b>0 (magnitude of a vector is always a positive quantity).
    Hence
    cosθ<0\cos\theta<0
    Or
    θ(π2,3π2)\theta\in \left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right). Hence θ\theta is an obtuse angle.
  • Question 3
    1 / -0
    Let a=i+2j+k,b=ij+k\vec {a} = i + 2j + k, \vec {b} = i - j + k and c=i+jk\vec {c} = i + j - k, a vector in the plane a\vec {a} and b\vec {b} whose projection on c\vec {c} is 13\dfrac {1}{\sqrt {3}} is _____
    Solution
    A vector in the plane of a\vec{a} and b\vec{b} would be ma+nbm\vec{a} + n\vec{b}
    i.e. (m+n)i+(2mn)j+(m+n)k(m + n)i + (2m - n)j + (m + n)k
    Its projection on c\vec{c} will be (ma+nb).cc\cfrac{(m\vec{a} + n\vec{b}).\vec{c}}{|c|}
    13=[(m+n)i+(2mn)j+(m+n)k].(i+jk)3\therefore \cfrac{1}{\sqrt{3}} = \cfrac{[(m + n)i + (2m - n)j + (m + n)k].(i + j - k)}{\sqrt{3}}
    1=m+n+2mnmn\Rightarrow 1 = m + n + 2m - n - m - n
    1=2mn\Rightarrow 1 = 2m - n 
    This relation is satisfied by m=1,n=1m = 1, n = 1
    Thus, the required vector can be 2i+j+2k2i + j + 2k
  • Question 4
    1 / -0
    Point PP is the endpoint of vector OPOP and point QQ is the endpoint of vector OQOQ as shown in the above figure. When the vectors OP\overline {OP} and OQ\overline {OQ} are added, calculate the length of the resultant vector.

    Solution
    Given in figure, P(3,7)P (-3,7) and Q(4,7)Q(4,-7) and point O(0,0)O (0,0)
    Then vector PO=PO= PO=(0,0)(3,7)=(3,7)\overline{PO}=(0,0)-(3,-7)=(3,-7)
    And vector OQ=OQ = OQ=(0,0)(4,5)=(4,5)\overline{OQ}=(0,0)-(4,-5)=(-4,5)
    Then length of vector PQ=PQ = PQ=(4+3)2+(57)2=1+45=2.24\overline{PQ}=\sqrt{(-4+3)^{2}+(5-7)^{2}}=\sqrt{1+4}\sqrt{5}=2.24
  • Question 5
    1 / -0
    The value of xx if x(i^+j^+k^)x(\hat { i } +\hat { j } +\hat { k } ) is a unit vector is
    Solution
    If x(i^+j^+k^)x(\hat i+\hat j+\hat k) is an unit vector then 
    x(i^+j^+k^)=1|x(\hat i+\hat j+\hat k)|=1
    x12+12+12=1\Rightarrow \bigg|x\sqrt{1^2+1^2+1^2} \bigg|=1
    3x=1\Rightarrow \sqrt 3|x|=1
    x=13\Rightarrow |x|=\dfrac{1}{\sqrt 3}
    x=±13\therefore x=\pm \dfrac{1}{\sqrt 3}
  • Question 6
    1 / -0
    If aandb\displaystyle \vec{a}\:and\:\vec{b} are the vectors ABandBC\displaystyle \overrightarrow{AB}\:and\:\overrightarrow{BC} determined by the adjacent sides of a regular hexagon. What are the vectors determined by the side CD and DE taken in order?
    Solution
    If a\overrightarrow { a } and b\overrightarrow { b } are the vectors determined by the adjacent sides of a regular hexagon.
    Let ABCDEFABCDEF be a regular hexagon such that 
    AB=a,BC=b\overrightarrow { AB } =\overrightarrow { a } ,\quad \overrightarrow { BC } =\overrightarrow { b }
    Now AC=AB+BC=a+bAD=2BC[AD=2BCandADBC]AD=2b\overrightarrow { AC } =\overrightarrow { AB } +\overrightarrow { BC } =\overrightarrow { a } +\overrightarrow { b } \\ \overrightarrow { AD } =2\overrightarrow { BC } \quad [\because \quad AD=2BC\quad and\quad AD\parallel BC]\\ \therefore \overrightarrow { AD } =2\overrightarrow { b }
    In ΔADC,AC+CD=ADCD=ADAC=2b(a+b )=baDE=AB=a\\ \Delta ADC,\overrightarrow { AC } +\overrightarrow { CD } =\overrightarrow { AD } \\ \overrightarrow { CD } =\overrightarrow { AD } -\overrightarrow { AC } =2\overrightarrow { b } -\left( \overrightarrow { a } +\overrightarrow { b }  \right) \\ \quad \quad \quad =\overrightarrow { b } -\overrightarrow { a } \\ \overrightarrow { DE } =-\overrightarrow { AB } =-\overrightarrow { a } \\ \quad \quad
    Hence correct option is AA 
    ba\overrightarrow { b } -\overrightarrow { a } ,a-\overrightarrow { a } \\ \quad \quad

  • Question 7
    1 / -0
    If pp , qq and rr are three non-coplanar vectors such that p+q+r =αsp + q + r  = \alpha s and q+r+s=βpq + r + s = \beta p . then p+q+r+s p +q +r +s  is equal to
    Solution
    Given 
    p+q+r=αsp+q+r=\alpha s----(1)
    q+r+s=βpq+r+s=\beta p----(2)

    p+q+r+sp+q+r+s =αs+s=\alpha s+s( from 1)
    =(α+1)s=(\alpha+1)s

  • Question 8
    1 / -0
    ABCDEFABCDEF is a regular hexagon whose centre is OO. The AB+AC+AD+AE+AF\overline { AB } +\overline { AC } +\overline { AD } +\overline { AE } +\overline { AF } is
    Solution
    Since it is a regular hexagon,
    AB=ED,AF=CD\overline { AB } =\overline { ED } ,\overline { AF } =\overline { CD }

    Now, AB+AC+AD+AE+AF=AE+ED+AC+CD+AD=3AD\overline { AB } +\overline { AC } +\overline { AD } +\overline { AE } +\overline { AF } =\overline { AE } +\overline { ED } +\overline { AC } +\overline { CD } +\overline { AD } =3\overline { AD }

    Now, since OO is the centre of the hexagon, OO will be the midpoint of ADAD
    3AD=6AO\therefore 3\overline{AD}=6\overline { AO }

  • Question 9
    1 / -0
    Two a\vec a and b\vec b are at an angle of 60o60^o with each other Their resultant makes an angle of 45o45^o with a\vec a If magnitude b=2|\vec b|= 2 units then a|\vec a| is:
    Solution
    The vectors make 60°60° with each other, and the resultant vector makes 45°45° with aa
    The other angle =15°=15°
    Using the side rule;
    a/sin15=2/sin45a=2sin15/sin45a=0.732a/\sin { 15 } =2/\sin { 45 } \\ a=2\sin { 15 } /\sin { 45 } \\ a=0.732
    Hence correct answer is (B)(B) 31\sqrt { 3 } -1

  • Question 10
    1 / -0
    The magnitude of resultant of three vectors of magnitude 1,21,2 and 33 whose direction are those of the sides of an equilateral triangle taken in order is:
    Solution
    The 11 magnitude vector in each cancels out to give only 11 and 22 magnitude vectors which are 120120 degrees apart, which gives a resultant of 3\sqrt{3} units
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