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  • Explain Solution R.D.Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 11 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 11 Maths Textbook Solution.

Answers (1)

write the ques number in proper format

Answer??

Hint :- When vectors are perpendicular their dot product is zero.

Given:-  \vec{a},\vec{b} & \vec{c}  be position vectors of A, B & C

\begin{aligned} &\therefore \overrightarrow{A B}=(\vec{b}-\vec{a}) \\ &\overrightarrow{B C}=(\vec{c}-\vec{b}) \\ &\overrightarrow{C A}=(\vec{a}-\vec{c}) \end{aligned}

If a vector given \left ( \vec{a}\times \vec{b} \right )+\left ( \vec{b}\times \vec{c} \right )+\left ( \vec{c}\times \vec{a} \right ) is perpendicular to the plane of triangle ABC then its dot product with the vector \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CA} must be zero

\begin{aligned} &\overrightarrow{A B} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})]\\ &=[(\vec{b}-\vec{a}) \cdot(\vec{a} \times \vec{b})]+[(\vec{b}-\vec{a}) \cdot(\vec{b} \times \vec{c})]+[(\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{a})]\\ \end{aligned}

\begin{aligned} &=[(\vec{b} \cdot(\vec{a} \times \vec{b})]-[\vec{a} \cdot(\vec{a} \times \vec{b})]+[\vec{b} \cdot(\vec{b} \times \vec{c})]-[\vec{a} \cdot(\vec{b} \times \vec{c})]+[(\vec{b} \cdot(\vec{c} \times \vec{a})]-[\vec{a} \cdot(\vec{c} \times \vec{a})]\\ \end{aligned}

=\left[\begin{array}{lll} \vec{b} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{b} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{b} & c \end{array}\right]+\left[\begin{array}{lll} \vec{b} & \vec{c} & a \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{c} & \vec{a} \end{array}\right]

\begin{aligned} &=0-0+0-[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{c}]-0 \\ &=0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because[\vec{b} \vec{c} \vec{a}]=[\vec{a} \vec{b} \vec{c}]) \end{aligned}

\begin{aligned} &\overrightarrow{B C} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \\\\ &=[\vec{c}-\vec{b}][(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \end{aligned}

=\left[\begin{array}{lll} \vec{c} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{c} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{b} & \vec{c} \end{array}\right]+\left[\begin{array}{lll} \vec{c} & \vec{c} & \vec{a} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{a} \end{array}\right]

    =\left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]+0+0-0-0-\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]

=0                                                                           =\left ( \because \left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]=\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ] \right )

     Similarly,

\begin{aligned} &\overrightarrow{C A} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \\\\ &=[\vec{a}-\vec{c}][(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \end{aligned}

=\left[\begin{array}{lll} \vec{a} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{b} & \vec{c} \end{array}\right]+\left[\begin{array}{lll} \vec{a} \vec{c} & \vec{a} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{a} \end{array}\right]

=0-\left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]\left [ \vec{c}\: \vec{a}\: \vec{b} \ \right ]-0+0-0

=0

As, dot product of all vectors is zero. We can say that \left ( \vec{a}\times \vec{b} \right )+\left ( \vec{b}\times \vec{c} \right )+\left ( \vec{c}\times \vec{a} \right ) is perpendicular to the plane of triangle ABC.

 

 

 

 

 

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