Find the vector equation of the plane passing through three points with position vectors
and
Also, find the coordinates of the point of intersection of this plane and the line
[CBSE 2013]
Let A, B and C be three point with position vector
and ![]()
Thus, ![]()
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As we know that cross product of two vectors gives a perpendicular vector so

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So, the equation of the required plane is
![]()
![]()
![]()
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Also we have to find the coordinates of the point of intersection of this plane and the line ![]()
Any point on the line
is of the form, p(3 + 2
, – 1 – 2
, – 1 +
)
Point p(3 + 2
, – 1 – 2
, – 1 +
) lies in the plane,
so,
9(3 + 2
) – 3(– 1 – 2
) – (– 1 +
) = 14
27 + 18
– 3 – 6
+ 1 –
= 14
11
= – 11
= – 1
Thus the required point of intersection is
p(3 + 2
, – 1 – 2
, – 1 +
)
put value of
in this equation
p[3 + 2(– 1), – 1 – 2(– 1), – 1 + (– 1)]
p(1, 1, – 2)
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[CBSE 2013]