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  • Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 6

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 6

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Answer:  \vec{r}=\lambda(\hat{i}+2 \hat{j}+3 \hat{k})

Hint: We use properties of vector

Given:  \vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=3

Solution: The requires line is perpendicular to the plane  \vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=3

Therefore, it is parallel to the normal  \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

Thus, the required line passes through the position vector  \overrightarrow{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}  and parallel to the vector \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

So, its vector equation is

\begin{aligned} &\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \vec{n} \\ &\vec{r}=(0 \hat{i}+0 \hat{\mathrm{j}}+0 \hat{k})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ &\vec{r}=\lambda(\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \end{aligned}

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