A line has direction d = 3i - k and passes through the point (8, 1, -1). A plane has equation r · n = 5, where n = -i + 10j - 3k. Which pair of checks shows that the line lies in the plane?

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Paper 1 · November 2010 · Vectors

A line has direction $d = 3i - k$ and passes through the point $(8, 1, -1)$ . A plane has equation $r \cdot n = 5$ , where $n = -i + 10j - 3k$ . Which pair of checks shows that the line lies in the plane?

Ad.n = 0 and the plane passes through the origin

Bd x n = 0 and the point (8, 1, -1) is in the plane

Cthe point (8, 1, -1) is in the plane, on its own

Dd.n = 0 and the point (8, 1, -1) is in the plane

Explanation

d\cdot n=3(-1)+0(10)+(-1)(-3)=0

, so the line is parallel to the plane. The point gives

8(-1)+1(10)+(-1)(-3)=-8+10+3=5

, so it satisfies the plane equation. A line parallel to a plane that shares one point with it lies wholly in it. Either check alone is not enough, and

d\times n=0

would mean the line is perpendicular to the plane, not in it.

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q14

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A line has direction d=3i−kd = 3i - kd=3i−k and passes through the point (8,1,−1)(8, 1, -1)(8,1,−1). A plane has equation r⋅n=5r \cdot n = 5r⋅n=5, where n=−i+10j−3kn = -i + 10j - 3kn=−i+10j−3k. Which pair of checks shows that the line lies in the plane?

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A line has direction $d = 3i - k$ and passes through the point $(8, 1, -1)$ . A plane has equation $r \cdot n = 5$ , where $n = -i + 10j - 3k$ . Which pair of checks shows that the line lies in the plane?