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CIE 9231 2024 June Paper 23 Q8

A Level / CIE / FP2

CIE 9231 2024 June Paper 23 Paper · Question 8

题目

Problem

The planes Π1\Pi_1 and Π2\Pi_2 do not intersect and are both perpendicular to i+2j+3k\mathbf{i}+2\mathbf{j}+3\mathbf{k}. The line ll intersects Π1\Pi_1 at the point (1,6,0)(1,6,0) and intersects Π2\Pi_2 at the point (3,6,0)(3,-6,0).

(a) Find Cartesian equations of Π1\Pi_1 and Π2\Pi_2.

[3]

(b) Express the vector equation of ll in the form

(xyz)=a+λb,\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{a}+\lambda\mathbf{b},

where a\mathbf{a} and b\mathbf{b} are vectors to be determined, and hence show that for points on ll,

12x+112y=1andz=0.\frac{1}{2}x+\frac{1}{12}y=1\qquad\text{and}\qquad z=0.
[2]
题目中文翻译

平面 Π1\Pi_1Π2\Pi_2 不相交,且都垂直于 i+2j+3k\mathbf{i}+2\mathbf{j}+3\mathbf{k}。直线 llΠ1\Pi_1 相交于点 (1,6,0)(1,6,0),并与 Π2\Pi_2 相交于点 (3,6,0)(3,-6,0)

(a) 求 Π1\Pi_1Π2\Pi_2 的笛卡尔方程。

(b) 将 ll 的向量方程写成

(xyz)=a+λb\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{a}+\lambda\mathbf{b}

的形式,其中 a\mathbf{a}b\mathbf{b} 为待确定的向量,并据此证明直线 ll 上的点满足

12x+112y=1以及z=0\frac{1}{2}x+\frac{1}{12}y=1\qquad\text{以及}\qquad z=0

解答