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The value of a, for which the points A,B,C with position  vectors 2\hat{i}-\hat{j}+\hat{k},\; \; \hat{i}-3\hat{j}-5\hat{k}\; \; and\; \; a\hat{i}-3\hat{j}+\hat{k}  respectively are the vertices of a right angled triangle at c are

  • Option 1)

    2 and 1

  • Option 2)

    –2 and –1

  • Option 3)

    –2 and 1

  • Option 4)

    2 and –1

 

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G gaurav
Answered Jul 07, 2018

As we have learned

Scalar Product of two vectors -

\vec{a}.\vec{b}> 0 \:an\: acute\: angle

\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle

\vec{a}.\vec{b}= 0 \:a\:right\: angle

- wherein

\Theta  is the angle between the vectors \vec{a}\:and\:\vec{b}

 

 

Position Vector -

If \vec{a} and \vec{b} are the position of vectors of two points A and B then

 \overrightarrow{AB}= \vec{b}-\vec{a}

\overrightarrow{AB}= P \vee of B - P\vee of A          

 

- wherein

 

 

\vec{AC}\cdot \vec{BC} = 0

\Rightarrow ((a\hat{i}-3\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+\hat{k}))\cdot ((a\hat{i}-3\hat{j}+\hat{k})- (\hat{i}-3\hat{j}-5\hat{k}))=0

( a-2 )( a-1 )= 0 

a = 1,2 

 

 

 


Option 1)

2 and 1

Option 2)

–2 and –1

Option 3)

–2 and 1

Option 4)

2 and –1

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

 

View All (1) Answers

G gaurav
Answered Jul 07, 2018

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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