The equation of the plane passing through the line of intersection of the planes \(\vec{r}.(\hat{i}+\hat{j}+\hat{k})\) = 1 and \(\vec{r}.(2\hat{i}+3\hat{j}-\hat{k})\)+ 4 = 0 and parallel to the x-axis is:
(a) \(\vec{r}.(\hat{j}-3\hat{k})\) + 6 = 0
(b) \(\vec{r}.(\hat{i}+3\hat{k})\) + 6 = 0
(c) \(\vec{r}.(\hat{i}-3\hat{k})\) + 6 = 0
(d) \(\vec{r}.(\hat{j}- 3\hat{k})\) - 6 = 0
Correct answer: (a) \(\vec{r}.(\hat{j}-3\hat{k})\) + 6 = 0
Explanation:
\(\vec{r}.(\hat{i}+ \hat{j} + \hat{k})\) - 1 = 0
⇒ x + y + z - 1 = 0
and \(\vec{r}.(2\hat{i}+ 3\hat{j} - \hat{k})\) + 4 = 0
⇒ 2x + 3y - z + 4 = 0
equation of planes through line of intersection of these planes is
(x + y + z – 1) + λ ( 2x + 3y – z + 4) = 0
⇒ (1 + 2λ) x + (1 + 3λ) y + (1– λ) z – 1 + 4λ = 0
But this plane is parallel to x–axis whose direction are (1, 0, 0)
∴ (1 + 2λ)1 + (1 + 3λ) 0 + (1 – λ) 0 = 0
λ = -\(\frac{1}{2}\)
∴ Required plane is
0 x + \((1 - \frac{3}{2})y\) + \((1 + \frac{1}{2})z\) - 1 + 4 \(( \frac{1}{2})\) = 0
⇒ \(\frac{-y}{2}\) + \(\frac{3}{2}\)z - 3 = 0
⇒ y - 3z + 6 = 0
⇒ \(\vec{r}.(\hat{j}-3\hat{k})\) + 6 = 0