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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 $$w$$ with the initial point $$(4,1,2)$$ and final point $$(1,6,5)$$.
    Solution
    Given points 
    $$A(4,1,2)$$ and $$B(1,6,5)$$
    $$\vec{OA}=4\hat{i}+\hat{j}+2\hat{k}$$
    $$\vec{OB}=\hat{i}+6\hat{j}+5\hat{k}$$
    vector 
    $$\vec{AB}=\vec{OB}-\vec{OA}$$
    $$\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-(4\hat{i}+\hat{j}+2\hat{k})$$
    $$\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-4\hat{i}-\hat{j}-2\hat{k}$$
    $$\vec{AB}=-3\hat{i}+5\hat{j}+3\hat{k}$$
    $$AB=(-3,5,3)$$
  • Question 2
    1 / -0
    Find $$2v$$, when $$u=(3,4,-2)$$ and $$v=(0,-4,0)$$.
    Solution
    Given points 
    $$u(3,4,-2)$$ and $$v(0,-4,0)$$
    $$\vec{u}=3\hat{i}+4\hat{j}-2\hat{k}$$
    $$\vec{v}=-4\hat{j}$$
    vector 
    $$2\vec{v}=2\times -4\vec{j}$$
    $$2\vec{v}=-8\vec{j}=(0,-8,0)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  • Question 10
    1 / -0
    The coordinates of a point in $$3$$-D space is $$(3,1,2)$$. 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 $$(3,1,2)$$ in $$3$$-D space.
    $$x,y$$ and $$z$$ coordinates are represented by vectors $$i,j$$ and $$k$$ respectively.
    Hence, position vector is given by $$3\hat i+\hat j+2\hat k$$.
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