Find the vector equation of line passing through a point $(1, 2, 3)$ and perpendicular to the plane $\to r (^i + 2^j - 5^k) + 9 = 0$ .

Open in App

Solution

The vector equation of a line passing through a point with position vector $\to a$ and parallel to $\to b$ is $\to r = \to a + λ \to b$
Given, the line passes through $(1, 2, 3)$
So, $\to a =^i + 2^j + 3^k$
Finding normal of plane
$\to r . (^i + 2^j - 5^k) + 9 = 0$
$\Rightarrow \to r . (^i + 2^j - 5^k) = - 9$
$\Rightarrow - \to r . (^i + 2^j - 5^k) = 9$
$\Rightarrow \to r . (-^i - 2^j + 5^k) = 9$
Comparing with $\to r . \to n = d$
$\to n = -^i - 2^j + 5^k$
since line is perpendicular to the plane, the line will be parallel to the normal of the plane
$∴ \to b = \to n = -^i - 2^j + 5^k$
Hence , $\to r = (^i + 2^j + 3^k) - λ (^i + 2^j - 5^k)$
$∴$ Vector equation of line is $\to r = (^i + 2^j + 3^k) - λ (^i + 2^j - 5^k)$

Suggest Corrections

Similar questions

Find the vector equation of the straight line passing through (1, 2, 3) and perpendicular to the plane $r . (^i + 2^j - 5^k) + 9 = 0$

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane

(1, 2, 3)

\to r . (^i + 2^j - 5^k) + 9 = 0

\vec{r} \cdot (\hat{i} + 2 \hat{j} - 5 \hat{k}) + 9 = 0 .

\vec{r} \cdot (\hat{i} + 2 \hat{j} - 5 \hat{k}) + 9 = 0 .

Join BYJU'S Learning Program