Question

Show that the points A,B,C with position vectors 2^i−^j+^k,^i−3^j−5^k and 3^i−4^j−4^k respectively, are the vertices of a right-angled triangle. Hence, find the area of the triangle.

Answer 1

To show that points A, B, and C form a right-angled triangle, we need to verify the dot product of two sides of the triangle is zero. The dot product is not zero, so the triangle is not right-angled. The area of the triangle is 3 units squared.

Step-by-step explanation:

To show that points A, B, and C form a right-angled triangle, we need to verify if the dot product of two sides of the triangle is equal to zero. We can find the vectors AB and AC by subtracting the position vectors. AB = (-1i - 2j - 6k) and AC = (-1i - 3j - 5k). The dot product of AB and AC is (-1 * -1) + (-2 * -3) + (-6 * -5) = 1 + 6 + 30 = 37.

Since the dot product is not zero, the triangle is not right-angled.

To find the area of the triangle, we can use the cross product of AB and AC. The cross product of two vectors gives a vector perpendicular to both vectors. The magnitude of this vector is equal to the area of the parallelogram formed by AB and AC, which is half the area of the triangle. The cross product of AB and AC is (4i - 4j + 2k). The magnitude of this vector is sqrt(4^2 + (-4)^2 + 2^2) = sqrt(36) = 6. So the area of the triangle is 6/2 = 3 units squared.