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in 3D Coordinate Geometry by (10.6k points)

सरल रेखा \(\frac{x-3}{-2}=\frac{y-4}{-5}=\frac{x+7}{13}\) का सदिश रूप होगा

(a) (3i + 4j – 7k) + λ( – 2i – 5j + 13k)

(b) ( – 2 – 5j + 13k) + λ(3i + 4j – 7k)

(c) ( – 3i – 4j + 7k) + λ( – 2i – j + 13k)

(d) इनमें से कोई नहीं ।

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  रेखा \(\frac{x-3}{-2}=\frac{y-4}{-5}=\frac{x+7}{13}\) 

बिंदु A(3, 4, – 7) से गुजरती है।

अतः बिंदु A का स्थिति सदिश
\(\overrightarrow{a}\) = 3i + 4j – 7k

दी हुई रेखा के दिक्-अनुपात – 2, – 5, 13 है।

∴ \(\overrightarrow{b}\) = – 2i – 5j + 13k

अत: अभीष्ट रेखा का समीकरण

\(\overrightarrow{r}\)  = \(\overrightarrow{a}\) + λ.\(\overrightarrow{b}\)

= (3i + 4j – 7k) + λ( – 2i – 5j + 13k)

अतः उत्तर का सही विकल्प (a) है।

 

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