Advertisements
Advertisements
Question
If points A, B and C have position vectors `2hati, hatj` and `2hatk` respectively, then show that ΔABC is an isosceles triangle.
Advertisements
Solution
Using distance formula
AB = `sqrt((0 - 2)^2 + (1 - 0)^2 + (0 - 0)^2`
= `sqrt(4 + 1)`
AB = `sqrt(5)` units
BC = `sqrt((0 - 0)^2 + (0 - 1)^2 + (2 - 0)^2`
= `sqrt(1 + 4)`
= `sqrt(5)`
BC = `sqrt(5)` units
AC = `sqrt((0 - 2)^2 + (0 - 0)^2 + (2 - 0)^2`
= `sqrt(4 + 4)`
= `2sqrt(2)`
∵ AB = BC = `sqrt(5) ≠ AC`
`\implies` ΔABC is an isosceles triangle.
APPEARS IN
RELATED QUESTIONS
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector
`2hati+3hatj+4hatk` to the plane `vecr` . `(2hati+hatj+3hatk)−26=0` . Also find image of P in the plane.
`veca and -veca` are collinear.
Two collinear vectors having the same magnitude are equal.
Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1) directed from A to B.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are `hati + 2hatj - hatk` and `-hati + hatj + hatk` respectively, externally in the ratio 2:1.
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of the x-axis.
Find the value of x for which `x(hati + hatj + hatk)` is a unit vector.
If θ is the angle between two vectors `veca` and `vecb`, then `veca . vecb >= 0` only when ______.
Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \text{ and } \vec{b} = 4\hat{i} + 4 \hat{j} - 2\hat{k}\]
Dot products of a vector with vectors \[\hat{i} - \hat{j} + \hat{k} , 2\hat{ i} + \hat{j} - 3\hat{k} \text{ and } \text{i} + \hat{j} + \hat{k}\] are respectively 4, 0 and 2. Find the vector.
If \[\hat{ a } \text{ and } \hat{b }\] are unit vectors inclined at an angle θ, prove that
\[\tan\frac{\theta}{2} = \frac{\left| \hat{a} -\hat{b} \right|}{\left| \hat{a} + \hat{b} \right|}\]
If \[\left| \vec{a} + \vec{b} \right| = 60, \left| \vec{a} - \vec{b} \right| = 40 \text{ and } \left| \vec{b} \right| = 46, \text{ find } \left| \vec{a} \right|\]
If \[\vec{p} = 5 \hat{i} + \lambda \hat{j} - 3 \hat{k} \text{ and } \vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k} ,\] then find the value of λ, so that \[\vec{p} + \vec{q}\] and \[\vec{p} - \vec{q}\] are perpendicular vectors.
If \[\vec{\alpha} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ and } \vec{\beta} = 2 \hat{i} + \hat{j} - 4 \hat{k} ,\] then express \[\vec{\beta}\] in the form of \[\vec{\beta} = \vec{\beta_1} + \vec{\beta_2} ,\] where \[\vec{\beta_1}\] is parallel to \[\vec{\alpha} \text{ and } \vec{\beta_2}\] is perpendicular to \[\vec{\alpha}\]
Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).
Show that the points whose position vectors are \[\vec{a} = 4 \hat{i} - 3 \hat{j} + \hat{k} , \vec{b} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} , \vec{c} = \hat{i} - \hat{j}\] form a right triangle.
If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?
If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.
If \[\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d}\] , show that \[\vec{a} - \vec{d}\] is parallel to \[\vec{b} - \vec{c}\] where \[\vec{a} \neq \vec{d} \text { and } \vec{b} \neq \vec{c}\] .
Find the sine of the angle between the vectors `vec"a" = 3hat"i" + hat"j" + 2hat"k"` and `vec"b" = 2hat"i" - 2hat"j" + 4hat"k"`.
The unit normal to the plane 2x + y + 2z = 6 can be expressed in the vector form as
Area of rectangle having vertices A, B, C and D will position vector `(- hati + 1/2hatj + 4hatk), (hati + 1/2hatj + 4hatk) (hati - 1/2hatj + 4hatk)` and `(-hati - 1/2hatj + 4hatk)` is
A line l passes through point (– 1, 3, – 2) and is perpendicular to both the lines `x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`. Find the vector equation of the line l. Hence, obtain its distance from the origin.
