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Need explanation for: - Vector Algebra - JEE Main-5

Let a,b \; and\, \; c be distinct non­-negative numbers. If the vectors a\hat{i}+a\hat{j}+c\hat{k},\; \hat{i}+\hat{k}\; and\; c\hat{i}+c\hat{j}+b\hat{k}  lie in a plane, then c is

  • Option 1)

    the arithmetic mean of a\; \; and\; \; b

  • Option 2)

    the geometric mean of  a\; \; and\; \; b

  • Option 3)

    the harmonic mean of a\; \; and\; \; b

  • Option 4)

    equal to zero

 
Answers (1)
83 Views
G gaurav

As we have learned

Scalar Triple Product -

\left [ \vec{a}\;\vec{b}\; \vec{c} \right ]

=\left (\vec{a}\times \vec{b}\right)\cdot \vec{c}= \vec{a}\cdot \left ( \vec{b} \times \vec{c}\right )

=\left (\vec{b}\times \vec{c}\right)\cdot \vec{a}= \vec{b}\cdot \left ( \vec{c} \times \vec{a}\right )

=\left (\vec{c}\times \vec{a}\right)\cdot \vec{b}= \vec{c}\cdot \left ( \vec{a} \times \vec{b}\right )

- wherein

Scalar Triple Product of three vectors \hat{a},\hat{b},\hat{c}.

 

 

[ a\hat{i}+a\hat{j}+c\hat{k}\; \;\; \; \; \; \hat{i}+\hat{k}\; \; \; \; \; \; c\hat{i}+c\hat{j}+b\hat{k}] = 0

\begin{vmatrix} a & a &c \\ 1& 0 & 1\\ c&c & b \end{vmatrix} = 0

- ac -a (b-c)+c^2 = 0 \\ ab = c^2

 

 

 

 

 

 

 


Option 1)

the arithmetic mean of a\; \; and\; \; b

Option 2)

the geometric mean of  a\; \; and\; \; b

Option 3)

the harmonic mean of a\; \; and\; \; b

Option 4)

equal to zero

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