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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    The distance between two points P and Q is d and the length of their projections of PQ on the co-ordinate planes are \[{{d}_{1}},{{d}_{2}},{{d}_{3}}\]. Then \[d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=k{{d}^{2}}\] where . k- is

    A) 1

    B) 5

    C) 3

    D) 2     

    Correct Answer: D

    Solution :

    • Here,        
    • \[{{d}_{1}}=d\cos ({{90}^{o}}-\alpha )\]                                                
    • \[{{d}_{2}}=d\cos ({{90}^{o}}-\beta )\]                                                
    • \[{{d}_{3}}=d\cos ({{90}^{o}}-\gamma )\]                                
    • Þ \[d_{1}^{2}+d_{2}^{2}+d_{3}^{2}={{d}^{2}}({{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma )\]                                                              
    • Þ \[d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=2{{d}^{2}}\];  \ \[k=2\].


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