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  • Provide solution for RD Sharma Maths Class 12 Chapter 25 Scale Triple Product Exercise Very Short Answer Question, question 4.

Provide solution for RD Sharma Maths Class 12 Chapter 25 Scale Triple Product Exercise Very Short Answer Question, question 4.

Answers (1)

Answer:

a=1, \frac{1}{2}

Hint:

Use scalar triple product formula.

Given:

 

Vectors \vec{\alpha}=\hat{\imath}+2 \hat{\jmath}+\hat{k}, \vec{\beta}=a \hat{\imath}+\hat{\jmath}+2 \hat{k} \text { and } \vec{\gamma}=\hat{\imath}+2 \hat{\jmath}+a \hat{k} are co planar.

Solution:

We have

\left.\begin{array}{l} \vec{\alpha}=\hat{\imath}+2 \hat{\jmath}+\hat{k} \\ \vec{\beta}=a \hat{\imath}+\hat{\jmath}+2 \hat{k} \\ \vec{\gamma}=\hat{\imath}+2 \hat{\jmath}+a \hat{k} \end{array}\right\}                         Eq.(i)

We know if three vectors are co planar then their scalar triple product is zero.

So, A/Q

α,β and γ are co planar then

\begin{aligned} &{\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]=0} \\ &\Rightarrow \vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma})=0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because[\vec{a} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=\vec{a} \cdot(\vec{b} \times \vec{c})] \\ &\Rightarrow(\hat{i}+2 \hat{j}+\hat{k}) \cdot\{(a \hat{i}+\hat{j}+2 \hat{k}) \times(\hat{i}+2 \hat{j}+a \hat{k})\}=0 \; \; \; \; \; \quad[\text { Using }(i)] \end{aligned}

\Rightarrow(\hat{i}+2 \hat{j}+\hat{k}) \cdot\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ a & 1 & 2 \\ 1 & 2 & a \end{array}\right|=0

\Rightarrow(\hat{i}+2 \hat{j}+\hat{k}) \cdot\left\{\hat{i}(a-4)-\hat{j}\left(a^{2}-2\right)+\hat{k}(2 a-1)\right\}=0

\begin{aligned} \Rightarrow(a-4)(\hat{\imath} . \hat{\imath})-\left(a^{2}-2\right)(\hat{\imath} . \hat{\jmath})+(2 a-1)(\hat{\imath} . \hat{k})+2(a-4)(\hat{\jmath} \cdot \hat{\imath})-2\left(a^{2}-2\right)(\hat{\jmath} \cdot \hat{\jmath}) \\ +2(2 a-1)(\hat{\jmath} . \hat{k})+(a-4)(\hat{k} \cdot \hat{\imath})-\left(a^{2}-2\right)(\hat{k} \cdot \hat{\jmath})+(2 a-1)(\hat{k} \cdot \hat{k})=0 \end{aligned}

\begin{aligned} &\Rightarrow(a-4) \cdot 1-0+0+0-2\left(a^{2}-2\right) \cdot 1+0+0-0+(2 a-1) \cdot 1=0 \\ &{\left[\begin{array}{l} \because \hat{\imath} . \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} \cdot \hat{k}=1, \hat{\imath} . \hat{\jmath}=\hat{\jmath} \cdot \hat{\imath}=0 \\ \hat{\imath} . \hat{k}=\hat{k} \cdot \hat{\imath}=0, \hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\jmath}=0 \end{array}\right]} \end{aligned}

\begin{aligned} &\Rightarrow a-4-2 a^{2}+4+2 a-1=0 \\ &\Rightarrow-2 a^{2}+3 a-1=0 \\ &\Rightarrow 2 a^{2}-3 a-1=0 \\ &\Rightarrow 2 a^{2}-2 a-a-1=0 \\ &\Rightarrow 2 a(a-1)-1(a-1)=0 \\ &\Rightarrow(a-1)(2 a-1)=0 \\ &\Rightarrow a-1=0 \text { or } 2 a-1=0 \\ &\Rightarrow a=1 \text { or } a=\frac{1}{2} \end{aligned}

 

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