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Given position vectors a = 2i + j − 3k, b = 3i − 2j + k, and c = i + 3j − k. Find the vector AB x AC and the area of the triangle ABC.

Mathematics

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4 parts
a

-14i - 6j - k

b

12233 units2\frac{1}{2}\sqrt{233} \text{ units}^2

c

14x+6y+z=3114x + 6y + z = 31

d

cos⁡−1(9699)\cos^{-1}\left(\frac{9}{\sqrt{699}}\right)

Step-by-step solution

15 steps

Here's the solution to question 7:

Given position vectors a=2i+j−3ka = 2i + j - 3k, b=3i−2j+kb = 3i - 2j + k, and c=i+3j−kc = i + 3j - k.

Part a· 3 steps

Find the vector AB⃗×AC⃗\vec{AB} \times \vec{AC}.

  1. Find the vector AB⃗\vec{AB}.

    AB⃗=b−a=(3i−2j+k)−(2i+j−3k)\vec{AB} = b - a = (3i - 2j + k) - (2i + j - 3k) AB⃗=(3−2)i+(−2−1)j+(1−(−3))k\vec{AB} = (3-2)i + (-2-1)j + (1-(-3))k AB⃗=i−3j+4k\vec{AB} = i - 3j + 4k

  2. Find the vector AC⃗\vec{AC}.

    AC⃗=c−a=(i+3j−k)−(2i+j−3k)\vec{AC} = c - a = (i + 3j - k) - (2i + j - 3k) AC⃗=(1−2)i+(3−1)j+(−1−(−3))k\vec{AC} = (1-2)i + (3-1)j + (-1-(-3))k AC⃗=−i+2j+2k\vec{AC} = -i + 2j + 2k

  3. Calculate the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}.

    AB⃗×AC⃗=∣ijk1−34−122∣\vec{AB} \times \vec{AC} = \begin{vmatrix} i & j & k \\ 1 & -3 & 4 \\ -1 & 2 & 2 \end{vmatrix} =i((−3)(2)−(4)(2))−j((1)(2)−(4)(−1))+k((1)(2)−(−3)(−1))= i((-3)(2) - (4)(2)) - j((1)(2) - (4)(-1)) + k((1)(2) - (-3)(-1)) =i(−6−8)−j(2−(−4))+k(2−3)= i(-6 - 8) - j(2 - (-4)) + k(2 - 3) =i(−14)−j(6)+k(−1)= i(-14) - j(6) + k(-1) =−14i−6j−k= -14i - 6j - k

    The vector AB⃗×AC⃗\vec{AB} \times \vec{AC} is −14i−6j−k\boxed{-14i - 6j - k}.

Part b

12233 units2\frac{1}{2}\sqrt{233} \text{ units}^2

Hence, find the area of the triangle ABC.

  1. The area of triangle ABC is half the magnitude of the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}.

    Area=12∣AB⃗×AC⃗∣\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|

  2. Calculate the magnitude of AB⃗×AC⃗\vec{AB} \times \vec{AC}.

    ∣AB⃗×AC⃗∣=∣−14i−6j−k∣=(−14)2+(−6)2+(−1)2|\vec{AB} \times \vec{AC}| = |-14i - 6j - k| = \sqrt{(-14)^2 + (-6)^2 + (-1)^2} =196+36+1=233= \sqrt{196 + 36 + 1} = \sqrt{233}

  3. Calculate the area of the triangle.

    Area=12233\text{Area} = \frac{1}{2} \sqrt{233}

    The area of the triangle ABC is 12233 units2\boxed{\frac{1}{2}\sqrt{233} \text{ units}^2}.

Part c

14x+6y+z=3114x + 6y + z = 31

Find the Cartesian equation of the plane ABC.

  1. The normal vector to the plane is n⃗=AB⃗×AC⃗=−14i−6j−k\vec{n} = \vec{AB} \times \vec{AC} = -14i - 6j - k.

    The Cartesian equation of a plane is of the form Ax+By+Cz=DAx + By + Cz = D, where (A,B,C)(A, B, C) are the components of the normal vector.
    So, the equation is −14x−6y−z=D-14x - 6y - z = D.

  2. Use a point on the plane to find DD.

    We can use point A with position vector a=2i+j−3ka = 2i + j - 3k, so (x,y,z)=(2,1,−3)(x, y, z) = (2, 1, -3).

    −14(2)−6(1)−(−3)=D-14(2) - 6(1) - (-3) = D −28−6+3=D-28 - 6 + 3 = D −31=D-31 = D

  3. Write the Cartesian equation of the plane.

    −14x−6y−z=−31-14x - 6y - z = -31

    Multiplying by -1 to make the leading coefficient positive:

    14x+6y+z=3114x + 6y + z = 31

    The Cartesian equation of the plane ABC is 14x+6y+z=31\boxed{14x + 6y + z = 31}.

Part d

cos⁡−1(9699)\cos^{-1}\left(\frac{9}{\sqrt{699}}\right)

Find the acute angle between the plane ABC and the plane x−y+z=5x - y + z = 5.

  1. Identify the normal vector of the plane ABC.

    From part c), n1⃗=14i+6j+k\vec{n_1} = 14i + 6j + k.

  2. Identify the normal vector of the second plane, x−y+z=5x - y + z = 5.

    n2⃗=i−j+k\vec{n_2} = i - j + k.

  3. Calculate the magnitudes of the normal vectors.

    ∣n1⃗∣=142+62+12=196+36+1=233|\vec{n_1}| = \sqrt{14^2 + 6^2 + 1^2} = \sqrt{196 + 36 + 1} = \sqrt{233} ∣n2⃗∣=12+(−1)2+12=1+1+1=3|\vec{n_2}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}

  4. Calculate the dot product of the normal vectors.

    n1⃗⋅n2⃗=(14)(1)+(6)(−1)+(1)(1)\vec{n_1} \cdot \vec{n_2} = (14)(1) + (6)(-1) + (1)(1) =14−6+1=9= 14 - 6 + 1 = 9

  5. Use the formula for the angle θ\theta between two planes

    cos⁡θ=∣n1⃗⋅n2⃗∣∣n1⃗∣∣n2⃗∣\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}.

    cos⁡θ=∣9∣2333=9699\cos \theta = \frac{|9|}{\sqrt{233} \sqrt{3}} = \frac{9}{\sqrt{699}}

  6. Find the acute angle θ\theta.

    θ=cos⁡−1(9699)\theta = \cos^{-1}\left(\frac{9}{\sqrt{699}}\right)

    The acute angle between the planes is cos⁡−1(9699)\boxed{\cos^{-1}\left(\frac{9}{\sqrt{699}}\right)}.

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