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Question: Let $\overline a = \hat i + \hat j$ and $\overline b = 2\hat i - \hat k$, then the point of intersec...

Let a‾=i^+j^\overline a = \hat i + \hat j and b‾=2i^−k^\overline b = 2\hat i - \hat k, then the point of intersection of the lines r‾×a‾=b‾×a‾\overline r \times \overline a = \overline b \times \overline a and r‾×b‾=a‾×b‾\overline r \times \overline b = \overline a \times \overline b is:

A

(3,-1,1)

B

(3,1,-1)

C

(-3,1,1)

D

(-3,-1,-1)

Answer

(3,1,-1)

Explanation

Solution

The given equations for the lines are:

  1. r‾×a‾=b‾×a‾\overline r \times \overline a = \overline b \times \overline a
  2. r‾×b‾=a‾×b‾\overline r \times \overline b = \overline a \times \overline b

The first equation can be rewritten as (r‾−b‾)×a‾=0‾(\overline r - \overline b) \times \overline a = \overline 0. This implies that the vector r‾−b‾\overline r - \overline b is parallel to a‾\overline a. Thus, r‾−b‾=ka‾\overline r - \overline b = k\overline a for some scalar kk, which gives the parametric equation of the first line as r‾=b‾+ka‾\overline r = \overline b + k\overline a.

Similarly, the second equation can be rewritten as (r‾−a‾)×b‾=0‾(\overline r - \overline a) \times \overline b = \overline 0. This implies that the vector r‾−a‾\overline r - \overline a is parallel to b‾\overline b. Thus, r‾−a‾=mb‾\overline r - \overline a = m\overline b for some scalar mm, which gives the parametric equation of the second line as r‾=a‾+mb‾\overline r = \overline a + m\overline b.

We are given: a‾=i^+j^=(1,1,0)\overline a = \hat i + \hat j = (1, 1, 0) b‾=2i^−k^=(2,0,−1)\overline b = 2\hat i - \hat k = (2, 0, -1)

The parametric equations of the lines are: Line 1: r‾=(2,0,−1)+k(1,1,0)=(2+k,k,−1)\overline r = (2, 0, -1) + k(1, 1, 0) = (2+k, k, -1) Line 2: r‾=(1,1,0)+m(2,0,−1)=(1+2m,1,−m)\overline r = (1, 1, 0) + m(2, 0, -1) = (1+2m, 1, -m)

To find the point of intersection, we equate the components of r‾\overline r from both equations: 2+k=1+2m2+k = 1+2m (x-component) k=1k = 1 (y-component) −1=−m-1 = -m (z-component)

From the y-component, we get k=1k=1. From the z-component, we get m=1m=1.

Now, we check if these values satisfy the x-component equation: 2+k=2+1=32+k = 2+1 = 3 1+2m=1+2(1)=31+2m = 1+2(1) = 3 The x-components match, so the lines intersect.

Substitute the value of k=1k=1 into the equation for Line 1: r‾=(2+1,1,−1)=(3,1,−1)\overline r = (2+1, 1, -1) = (3, 1, -1)

Alternatively, substitute the value of m=1m=1 into the equation for Line 2: r‾=(1+2(1),1,−1)=(3,1,−1)\overline r = (1+2(1), 1, -1) = (3, 1, -1)

Both methods yield the same point of intersection. The point of intersection is (3,1,−1)(3, 1, -1).