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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 27 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 27 sub question (i) maths textbook solution

Answers (1)

Answer:

\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k})

Hint: To solve this we use determinant method

Given:

\begin{aligned} &\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} \\\\ &\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} \\\\ &\vec{c} \cdot \vec{d}=15 \end{aligned}

Solution:

D is perpendicular to a and b both. Hence, parallel to a*b

\begin{aligned} &\vec{a} \times \vec{b}=\vec{c} \\\\ &\vec{a} \times \vec{b}=\vec{d} \\\\ &\vec{d}=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{array}\right| \end{aligned}

\begin{aligned} &\vec{d}=\lambda(32 \hat{i}-\hat{j}-14 \hat{k}) \\\\ &\vec{c} \cdot \vec{d}=15 \\\\ &\lambda(2 \hat{i}-\hat{j}+4 \hat{k})(32 \hat{i}-\hat{j}-14 \hat{k})=15 \end{aligned}

\begin{aligned} &\lambda(64+1-56)=15 \\\\ &9 \lambda=15 \\\\ &\lambda=\frac{5}{3} \\\\ &\vec{d}=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \end{aligned}

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