Question
Download Solution PDFConsider three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k. If p, q and r denote the position vector of three non-collinear points, then the equation of the plane containing these points is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k.
Concept:
The equation of plane passing through three points a , b and c is
\(\rm (\vec r-a)\cdot((b-a)\times(c-a))=0\)
Calculation:
Three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k.
Now,
q - p = - i + j - 5k
and r - p = 0i + 0j - 3k
Then the normal vector
\( \rm \vec n=(q-p)\times(r-p)=\begin{bmatrix}i&j&k\\-1&1&-5\\0&0&-3\end{bmatrix}\)
\(\rm \vec n = -3i-3j=-3(i+j)\)
Then the equation of plane is
\(\rm (\vec r-\vec p)⋅ \vec n=0\)
put all the values then
((xi + yj + zk) - (2i + 3j + 4k))⋅ (-3(i+j)) = 0
((xi + yj + zk) - (2i + 3j + 4k))⋅ (i+j) = 0
(xi + yj + zk) ⋅ (i+j) - (2i + 3j + 4k) ⋅ (i+j) = 0
x + y - 2 - 3 = 0
x + y - 5 = 0
Hence the option (2) is correct.
Last updated on May 26, 2025
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