NCERT Class 12 · Ch 10 Miscellaneous Exercise · Q12
NCERT Class 12 Maths · Chapter 10 Vector Algebra · Miscellaneous Exercise Question 12
For a=(1,4,2), b=(3,−2,7), c=(2,−1,4), find d perpendicular to a,b with c·d=15.
💡 Concept Hints
- Write the vector components and identify the required direction or geometric relation.
- Use the magnitude, dot-product, cross-product or section formula that matches the requested quantity.
✅ Step-by-Step Solution
Final Answer: \(\left[\begin{matrix}\frac{160}{3}\\- \frac{5}{3}\\- \frac{70}{3}\end{matrix}\right]\)
- A perpendicular direction is a×b=\(\left[\begin{matrix}32\\-1\\-14\end{matrix}\right]\). Since a,b are independent, write d=t(a×b).
- c·(a×b)=\(9\), so t=\(\frac{5}{3}\).
- Thus d=\(\left[\begin{matrix}\frac{160}{3}\\- \frac{5}{3}\\- \frac{70}{3}\end{matrix}\right]\); direct dot products give a·d=b·d=0 and c·d=15.
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