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JEE Mains · Maths · STD 12 - 11. three dimension geometry

समतलों \(\overrightarrow{ r } \cdot(\hat{ i }+\hat{ j }+\hat{ k })=1\) तथा \(\overrightarrow{ r } \cdot(2 \hat{ i }+3 \hat{ j }-\hat{ k })+4=0\) की प्रतिच्छेदन रेखा से होकर जाने वाले तथा \(x\)-अक्ष के समांतर समतल का समीकरण है

  1. A \(\vec{r} \cdot(\hat{j}-3 \hat{k})+6=0\)
  2. B \(\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+3 \hat{\mathrm{k}})+6=0\)
  3. C \(\vec{r} \cdot(\hat{i}-3 \hat{k})+6=0\)
  4. D \(\vec{r} \cdot(\hat{j}-3 \hat{k})-6=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\vec{r} \cdot(\hat{j}-3 \hat{k})+6=0\)

Step-by-step Solution

Detailed explanation

Equation of planes are \(\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1=0 \Rightarrow \mathrm{x}+\mathrm{y}+\mathrm{z}-1=0\) and…
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