Find the vector and Cartesian equations of the plane passing through the points with position vectors $3 \to i + 4 \to j + 2 \to k, 2 \to i - 2 \to j - \to k$ and $7 \to i + \to k$ .

Open in App

Solution

Let the point $\to A$ be $(3, 4, 2)$ , point $\to B$ be $(2, - 2, - 1)$ and point $\to C$ be $(7, 0, 1)$
Which gives $\to A C = - 4 i + 4 j + k$ and $\to B C = - 5 i - 2 j - 2 k$
Therefore the directional ratios of normal of required plane is cross product of $\to A C$ and $\to B C$
$\to A C \times \to B C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 4 & 4 & 1 - 5 & - 2 & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (- 8 + 2) -^j (8 + 5) +^k (8 + 20) = - 6^i - 13^j + 28^k$
We get $- 6, - 13, 28$ as directional ratios of normal of required plane
Therefore the equation of plane is $- 6 x - 13 y + 28 z = k$ , where $k$ is constant
Now substitute one point on plane equation inorder to get the value of constant $k$
We get $- 18 - 52 + 56 = - 14 = k$ , Therefore the required equation is $- 6 x - 13 y + 28 z + 14 = 0$
The cartesian equation is $6 x + 13 y - 28 z - 14 = 0$ and the vector equation is $\to r . (6 i + 13 j - 28 k) = 14$

Suggest Corrections

Similar questions

3 \hat{i} + 4 \hat{j} + 2 \hat{k}, 2 \hat{i} - 2 \hat{j} - \hat{k} and 7 \hat{i} + 6 \hat{k} .

2 i + 4 j + 2 k

2 i + 3 j + 5 k

3 i - 2 j + k

\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}) .

2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}

3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k} and - \hat{i} + 4 \hat{j} - 2 \hat{k}

¯ A B

3 i + j + 2 k, i - 2 j - 4 k

Join BYJU'S Learning Program