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Mathematics Question on Three Dimensional Geometry

A vector perpendicular to the plane containing the points A(1,−1,2),B(2,0,−1),C(0,2,1)A (1, -1, 2), B (2, 0, -1), C (0,2,1) is

A

8i^+4j^+4k^ 8 \hat i + 4 \hat j + 4 \hat k

B

4i^+8j^−4k^ 4 \hat i + 8 \hat j - 4 \hat k

C

i^+j^−k^ \hat i + \hat j - \hat k

D

3i^+j^+2k^ 3 \hat i + \hat j + 2 \hat k

Answer

8i^+4j^+4k^ 8 \hat i + 4 \hat j + 4 \hat k

Explanation

Solution

We know that a vector perpendicular to the plane containing the points
A,B,CA , B , C is given by A×B+B×C+C×AA \times B + B \times C + C \times A
We have, A=i^−j^+2k^,B⃗=2i^+0j^−k^A = \hat{i} - \hat{j} + 2\hat{k}, \vec{B} = 2\hat{i} + 0 \hat{j} - \hat{k}
and C=0i^+2j^+k^C= 0 \hat{i} + 2 \hat{j} + \hat{k}
Now, A×B=∣i^j^k^ 1−12 20−1∣=i^+5j^+2k^A \times B= \begin{vmatrix}\hat{i}& \hat{j}&\hat{k}\\\ 1&-1&2\\\ 2&0&-1\end{vmatrix} = \hat{i} + 5\hat{j} +2 \hat{k}
B×C=∣i^j^k^ 20−1 021∣=2i^−2j^+4k^B \times C = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 2&0&-1\\\ 0&2&1\end{vmatrix} = 2\hat{i} - 2\hat{j} + 4\hat{k}
C×A⃗=∣i^j^k^ 021 1−12∣=5i^+j^−2k^C \times\vec{A} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 0&2&1\\\ 1&-1&2\end{vmatrix} = 5\hat{i} + \hat{j} - 2\hat{k}
Thus, A×B+B×C+C×B=(i^+5j^+2k^)A \times B + B \times C + C \times B =(\hat{ i }+5 \hat{ j }+2 \hat{ k })
+(2i^−2j^+4k^)+(5i^+j^−2k^)+(2 \hat{ i }-2 \hat{ j }+4 \hat{ k })+(5 \hat{ i }+\hat{ j }-2 \hat{ k })
=8i^+4j^+4k^=8 \hat{ i }+4 \hat{ j }+4 \hat{ k }