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Vector Algebra Test - 18

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Vector Algebra Test - 18
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  • Question 1
    1 / -0
    Find the vector ww with the initial point (4,1,2)(4,1,2) and final point (1,6,5)(1,6,5).
    Solution
    Given points 
    A(4,1,2)A(4,1,2) and B(1,6,5)B(1,6,5)
    OA=4i^+j^+2k^\vec{OA}=4\hat{i}+\hat{j}+2\hat{k}
    OB=i^+6j^+5k^\vec{OB}=\hat{i}+6\hat{j}+5\hat{k}
    vector 
    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    AB=i^+6j^+5k^(4i^+j^+2k^)\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-(4\hat{i}+\hat{j}+2\hat{k})
    AB=i^+6j^+5k^4i^j^2k^\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-4\hat{i}-\hat{j}-2\hat{k}
    AB=3i^+5j^+3k^\vec{AB}=-3\hat{i}+5\hat{j}+3\hat{k}
    AB=(3,5,3)AB=(-3,5,3)
  • Question 2
    1 / -0
    Find 2v2v, when u=(3,4,2)u=(3,4,-2) and v=(0,4,0)v=(0,-4,0).
    Solution
    Given points 
    u(3,4,2)u(3,4,-2) and v(0,4,0)v(0,-4,0)
    u=3i^+4j^2k^\vec{u}=3\hat{i}+4\hat{j}-2\hat{k}
    v=4j^\vec{v}=-4\hat{j}
    vector 
    2v=2×4j2\vec{v}=2\times -4\vec{j}
    2v=8j=(0,8,0)2\vec{v}=-8\vec{j}=(0,-8,0)

  • Question 3
    1 / -0
    Find the magnitude of the vector which joins the point AA(4,5,6)(4,5,6) to BB(10,11,12)(10,11,12).
    Solution
    Given points 
    A(4,5,6)A(4,5,6) and B(10,11,12)B(10,11,12)
    OA=4i^+5j^+6k^\vec{OA}=4\hat{i}+5\hat{j}+6\hat{k}
    OB=10i^+11j^+12k^\vec{OB}=10\hat{i}+11\hat{j}+12\hat{k}
    vector 
    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    AB=10i^+11j^+12k^(4i^+5j^+6k^)\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-(4\hat{i}+5\hat{j}+6\hat{k})
    AB=10i^+11j^+12k^4i^5j^6k^\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-4\hat{i}-5\hat{j}-6\hat{k}
    AB=6i^+6j^+6k^\vec{AB}=6\hat{i}+6\hat{j}+6\hat{k}
    magnitude of AB 
    AB=62+62+62\left | \vec{AB} \right |=\sqrt{6^2+6^2+6^2}

    AB=36+36+36\left | \vec{AB} \right |=\sqrt{36+36+36}

    AB=36×3\left | \vec{AB} \right |=\sqrt{36\times 3}

    AB=63\left | \vec{AB} \right |=6\sqrt{3}

    AB=6×1.732\left | \vec{AB} \right |=6\times 1.732

    AB=10.39\left | \vec{AB} \right |=10.39

  • Question 4
    1 / -0
    Which of the following is not a unit vector for all values of θ\theta?
    Solution
    Let nn be any vector.
    Then nn is a unit vector, if n=1||n||=1
    We know that, cos2θ+sin2θ=1\cos^2 \theta + \sin^2 \theta=1
    For n=(sin2θ)i(cosθ)jn=(\sin 2\theta)i - (\cos \theta)j
    n=sin22θ+cos2θ1||n||=\sin^{2} 2\theta + \cos^{2} \theta \neq 1
    Hence, (sin22θ)i+(cos2θ)j(\sin^{2} 2\theta)i + (\cos^{2} \theta)j is not a unit vector.
  • Question 5
    1 / -0
    Find the vector ww with the initial point (8,10,3)(8,-10,3) and final point (1,10,7)(1,10,7).
    Solution
    Given points 
    A(8,10,3)A(8,-10,3) and B(1,10,7)B(1,10,7)
    OA=8i^10j^+3k^\vec{OA}=8\hat{i}-10\hat{j}+3\hat{k}
    OB=i^+10j^+7k^\vec{OB}=\hat{i}+10\hat{j}+7\hat{k}
    vector w from A to B 
    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    w=i^+10j^+7k^(8i^10j^+3k^)\vec{w}=\hat{i}+10\hat{j}+7\hat{k}-(8\hat{i}-10\hat{j}+3\hat{k})
    w=i^+10j^+7k^8i^+10j^3k^\vec{w}=\hat{i}+10\hat{j}+7\hat{k}-8\hat{i}+10\hat{j}-3\hat{k}
    AB=7i^+20j^+4k^=(7,20,4)\vec{AB}=-7\hat{i}+20\hat{j}+4\hat{k}=(-7,20,4)
  • Question 6
    1 / -0
    The vector joining vector A\vec{A} to B\vec{B}  is represented by:
    Solution
    Given vectors 
    A\vec{A} and B\vec{B}
    AB=BA\vec{AB}=\vec{B}-\vec{A}

  • Question 7
    1 / -0
    Find the magnitude of the vector which joins the point AA(1,3,1)(-1,-3,-1) to BB(0,0,0)(0,0,0).
    Solution
    Given points 
    A(1,3,1)A(-1,-3,-1) and B(0,0,0)B(0,0,0)
    OA=i^3j^k^\vec{OA}=-\hat{i}-3\hat{j}-\hat{k}
    OB=0\vec{OB}=0
    vector AB  from A to B 
    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    AB=0(i^3j^k^)\vec{AB}=0-(-\hat{i}-3\hat{j}-\hat{k})
    AB=+i^+3j^+k^)\vec{AB}=+\hat{i}+3\hat{j}+\hat{k})
    magnitude of AB 
    AB=12+32+12\left | \vec{AB} \right |=\sqrt{1^2+3^2+1^2}

    AB=1+9+1\left | \vec{AB} \right |=\sqrt{1+9+1}

    AB=11\left | \vec{AB} \right |=\sqrt{11}

  • Question 8
    1 / -0
    Find the vector ww with the initial point (a,b,c)(a,b,c) and final point (a+1,b+2,c+3)(a+1,b+2,c+3).
    Solution
    Given points 
    A(a,b,c)A(a,b,c) and B(a+1,b+2,c+3)B(a+1,b+2,c+3)
    OA=ai^+bj^+ck^\vec{OA}=a\hat{i}+b\hat{j}+c\hat{k}
    OB=(a+1)i^+(b+2)j^+(c+3)j^\vec{OB}=(a+1)\hat{i}+(b+2)\hat{j}+(c+3)\hat{j}
    vector ww from AA to BB 
    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    w=(a+1)i^+(b+2)j^+(c+3)j^(ai^+bj^+ck^)\vec{w}=(a+1)\hat{i}+(b+2)\hat{j}+(c+3)\hat{j}-(a\hat{i}+b\hat{j}+c\hat{k})
    w=ai^+bj^+ck^+i^+2j^+3j^ai^bj^ck^\vec{w}=a\hat{i}+b\hat{j}+c\hat{k}+\hat{i}+2\hat{j}+3\hat{j}-a\hat{i}-b\hat{j}-c\hat{k}
    w=i^+2j^+3j^=(1,2,3)\vec{w}=\hat{i}+2\hat{j}+3\hat{j}=(1,2,3)

  • Question 9
    1 / -0
    Find the magnitude of the vector which joins the point AA(a,2,c)(a,2,c) to BB(a+1,5,c+3)(a+1,5,c+3).
    Solution
    Given points 
    A(a,2,c)A(a,2,c) and B(a+1,5,c+3)B(a+1,5,c+3)
    OA=ai^+2j^+ck^\vec{OA}=a\hat{i}+2\hat{j}+c\hat{k}
    OB=(a+1)i^+5j^+(c+3)j^\vec{OB}=(a+1)\hat{i}+5\hat{j}+(c+3)\hat{j}

    AB=OBOA\vec{AB}=\vec{OB}-\vec{OA}
    AB=(a+1)i^+5j^+(c+3)j^(ai^+2j^+ck^)\vec{AB}=(a+1)\hat{i}+5\hat{j}+(c+3)\hat{j}-(a\hat{i}+2\hat{j}+c\hat{k})
    AB=ai^+5j^+ck^+i^+3j^ai^2j^ck^\vec{AB}=a\hat{i}+5\hat{j}+c\hat{k}+\hat{i}+3\hat{j}-a\hat{i}-2\hat{j}-c\hat{k}
    AB=i^+3j^+3j^\vec{AB}=\hat{i}+3\hat{j}+3\hat{j}

    magnitude of AB 
    AB=12+32+32\left | \vec{AB} \right |=\sqrt{1^2+3^2+3^2}

    AB=1+9+9\left | \vec{AB} \right |=\sqrt{1+9+9}

    AB=19\left | \vec{AB} \right |=\sqrt{19}
    none of option is matched 

  • Question 10
    1 / -0
    The coordinates of a point in 33-D space is (3,1,2)(3,1,2). Then the position vector of the point is:
    Solution
    A position vector is a Euclidean vector that represents the position of a point P in space in relation to an arbitrary reference. 
    Given point is (3,1,2)(3,1,2) in 33-D space.
    x,yx,y and zz coordinates are represented by vectors i,ji,j and kk respectively.
    Hence, position vector is given by 3i^+j^+2k^3\hat i+\hat j+2\hat k.
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