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  • Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 28 maths

Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 28 maths

Answers (1)

Answer:

\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k})

Hint:

 To solve this we use determinant method

Given:

\begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}

Solution:   

\begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}

Let

\begin{aligned} &\vec{d}=\vec{a}+\vec{b} \\\\ &\vec{d}=4 \hat{i}+4 \hat{j}+0 \hat{k} \end{aligned}

And

\begin{aligned} &\vec{e}=\vec{a}-\vec{b} \\\\ &=2 \hat{i}+0 \hat{j}+4 \hat{k} \end{aligned}

Let  \vec{f} be any vector perpendicular to both  \vec{d} \; \& \; \vec{e}, hence parallel to \vec{d} \times \vec{e}

\begin{aligned} &\therefore \vec{f}=\vec{d} \times \vec{e} \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array}\right| \\\\ &=16 \hat{i}-16 \hat{j}-8 \hat{k} \end{aligned}

\begin{aligned} &=\frac{16\hat {i}-16 \hat{j}-8 \hat {k}}{\sqrt{(16)^{2}+(-16)^{2}+(-8)^{2}}} \\\\ &=\frac{16}{24} \hat{i}-\frac{16}{24} \hat{j}-\frac{8}{24} \hat{k} \\\\ &=\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k}) \end{aligned}

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