Consider a triangle PQR with O as the origin. A point S satisfies the condition OP·OQ + OR·OS = OR·OP + OQ·OS = OQ·OR + OP·OS. What is the role of point S in triangle PQR?

orthocenter
incenter
centroid
circumcenter

Correct answer: orthocenter

Solution

The point S satisfies the condition OP·OQ + OR·OS = OR·OP + OQ·OS = OQ·OR + OP·OS, which implies that S is the orthocenter of triangle PQR, as the altitudes of the triangle intersect at this point.

Related JEE Advanced Maths questions

Given that cos α ≠ 1, cos β ≠ 1, and cos γ ≠ 1, the vectors →a = i cos α + j cos β + k cos γ, →b = i + j cos β + k, and →c = i + j + k cos γ are
Given the vectors →a = a→i + 2→j − 3→k, →b = →i + 2a→j − 2→k, and →c = 2→i − a→j + →k, if (→a × →b) × (→b × →c) × (→c × →a) equals →0, what is the value of a?
In the triangle ΔPQR, the vectors →a, →b, and →c represent →QR, →RP, and →PQ respectively. If the magnitudes are |→a| = 12 and |→b| = 4√3, and the dot product →b ⋅ →c equals 24, which of the following statements is correct?
Consider three vectors →x, →y, and →z, each having a magnitude of √2, with the angle between any two of them being π/3. If →a is a non-zero vector orthogonal to both →y and →z, and →b is a non-zero vector orthogonal to both →x and →z, which of the following is true?
Let a, b, and c represent three unit vectors that are not in the same plane, with the angle between each pair being π/3. If the expression a × b + b × c equals pa + qb + rc, where p, q, and r are constants, what is the value of (p² + 2q² + r²)/q²?
The vector a = (1, 3, sin 2α) forms an angle greater than 90° with the z-axis. Given that 2α is negative, and vectors b and c are perpendicular, b ⋅ c equals zero. The equation tan² α − tan α − 6 = 0 has solutions tan α = 3 or −2. If tan α = 3, then sin 2α = 2 tan α / (1 + tan² α) = 3/5, which is positive and contradicts the condition (2α < 0). Thus, tan α = −2. For this value, sin 2α = 2 tan α / (1 − tan² α) = 4/3, which is greater than zero. However, sin 2α must be negative, placing 2α in the third quadrant and α/2 in the first quadrant. The square root of sin(α/2) is valid, and α is given by (4n + 1)π − tan⁻¹ 2.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →