Find the vector and certesian equations of the line passing through the points $(1, 2, 3)$ and parallel to the planes $¯ ¯ ¯ r (¯ i - ¯ j + 2 ¯ ¯ ¯ k) = 5$ and $¯ ¯ ¯ r (3 ¯ i - ¯ j + ¯ ¯ ¯ k) = 6$

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Solution

The vector equation of a line passing through a point with position vector $\overrightarrow a $ and parallel to a vector $\to a$ is
$\to r = \to a + λ \to b$
Given, the line passes through ( $1, 2, 3$ )
So, $\to a = 1 ˆ i + 2 ˆ j + 3 ˆ k$
Given, line is parallel to both planes
Line is perpendicular to normal of both planes.
i.e. $\to b$ is perpendicular to normal of both planes.
We know that
$\to a \times \to b$ is perpendicular to both $\to a a n d \to b$
So, $\to b$ is cross product of normals of planes $\to r \cdot (ˆ i - ˆ j + 2 ˆ k) = 5$ and $\to r \cdot (3 ˆ i - ˆ j + ˆ k) = 6$
Required normal
= $\begin{matrix} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 1 & - 1 & 2 3 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ˆ i (- 1 (1) + 1 (2)) - ˆ j (1 (1) - 3 (2)) + ˆ k (1 (- 1) - 3 (- 1)) = ˆ i (- 1 + 2) - ˆ j (1 - 6) + ˆ k (- 1 + 3) = 1 ˆ i - 5 ˆ j + 2 ˆ k \end{matrix}$
thus $\to b = 1 ˆ i - 5 ˆ j + 2 ˆ k$
Now putting value of $\to a a n d \to b$ in formula
$\to r = \to a + λ \to b$
$= (1^i + 2^j + 3^k) + λ (1^i - 5^j + 2^k)$
Therefore, the equation of the line is
$(1^i + 2^j + 3^k) + λ (1^i - 5^j + 2^k)$

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Similar questions

\vec{r} \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} + \hat{j} + \hat{k}) = 6 .

\vec{r} \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} + \hat{j} + 2 \hat{k}) = 6

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $r . (^i -^j + 2^k) = 5$ and $r . (3^i +^j + 2^k) = 6$ .

\hat{i} + \hat{j} - 2 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k} and \hat{i} + 2 \hat{j} + \hat{k} .

\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + λ (2 \hat{i} - 2 \hat{j} + \hat{k}) .

(1, 2, 3)

¯ r \cdot (¯ i - ¯ j + 2 ¯ k) = 5

¯ r \cdot (3 ¯ i + ¯ j + ¯ k) = 6

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