The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k. The line l_2 is parallel to the vector -2i + j + 3k and passes through the point B, whose position vector is -3i - j + 2k. The point P on l_1 and the point Q on l_2 are such that PQ is perpendicular to both l_1 and l_2. Find (i) the length PQ. (ii) the cartesian equation of the plane Π containing PQ and l_1. (iii) the perpendicular distance of A from Π.

Photo AI

The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k - CIE - A-Level Further Maths - Question 10 - 2014 - Paper 1

Question 10

The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k. The line l_2 is parallel to the vector -2i ... show full transcript

Worked Solution & Example Answer:The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k - CIE - A-Level Further Maths - Question 10 - 2014 - Paper 1

Step 1

the length PQ.

96%

114 rated

Answer

To find the length PQ, we first determine the direction vectors of the lines.

For line l_1, the direction vector is: $d_1 = i - 2j - 3k$

For line l_2, the direction vector is: $d_2 = -2i + j + 3k$

Next, we find points A and B:

A, with position vector: $A = (3, 3, -4)$
B, with position vector: $B = (-3, -1, 2)$

We compute the vector BA: $BA = A - B = (3 - (-3), 3 - (-1), -4 - 2) = (6, 4, -6)$

The shortest distance (d) between two skew lines is given by: $d = \frac{|(d_1 \times d_2) \cdot BA|}{|d_1 \times d_2|}$

We calculate:

The cross product (d_1 \times d_2).
The absolute value of the dot product ( |(d_1 \times d_2) \cdot BA| ).

Solving gives us:

$d = \frac{4}{\sqrt{3}} \approx 2.31$

Step 2

the cartesian equation of the plane Π containing PQ and l_1.

99%

104 rated

Answer

To find the cartesian equation of plane Π, we require a point on the plane and a normal vector.

We can take point A as a point on l_1, and we already have the direction vector of l_1. For the plane, since PQ is perpendicular to both l_1 and l_2, the normal vector can be calculated as:

$N = d_1 \times d_2$

Calculating gives us: $N = \begin{vmatrix} i & j & k \\ 1 & -2 & -3 \\ -2 & 1 & 3 \end{vmatrix}$

Thus, the equation of the plane can be expressed as: $Nx + My + Pz = D$ , where N, M, P are components of vector N, and D can be determined using point A.

Upon simplification, we find: $4x + 5y + z = 12$ .

Step 3

the perpendicular distance of A from Π.

96%

101 rated

Answer

The perpendicular distance from point A to the plane Π can be calculated using:

$distance = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Here, (x_1, y_1, z_1) are the coordinates of point A and (A, B, C) are the coefficients of the plane equation. Substituting gives us:

$distance = \frac{|4(3) + 5(3) + 1(-4) - 15|}{\sqrt{4^2 + 5^2 + 1^2}}$

This evaluates to: $\frac{38}{\sqrt{42}} \approx 5.86$ .

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k - CIE - A-Level Further Maths - Question 10 - 2014 - Paper 1

Worked Solution & Example Answer:The line l_1 is parallel to the vector i - 2j - 3k and passes through the point A, whose position vector is 3i + 3j - 4k - CIE - A-Level Further Maths - Question 10 - 2014 - Paper 1

the length PQ.

the cartesian equation of the plane Π containing PQ and l_1.

the perpendicular distance of A from Π.

Join the A-Level students using SimpleStudy...

Other A-Level Further Maths topics to explore