Exams › JEE Main › Maths
A sphere is given by the equation \(x^2+y^2+z^2-10z=0\). If one endpoint of a diameter is \((-3,4,5)\), then what are the coordinates of the opposite endpoint?
- \((-3,-4,-5)\)
- \((3,4,5)\)
- \((3,4,-5)\)
- \((-3,4,-5)\)
Correct answer: \((3,4,5)\)
Solution
The sphere \(x^2+y^2+z^2-10z=0\) has center \((0,0,5)\). Since one endpoint of a diameter is \((-3,4,5)\), the opposite endpoint is its reflection across the center, which is \((3,4,5)\).
Related JEE Main Maths questions
- A line makes equal angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive directions of the coordinate axes. If \(\theta\) satisfies \[ \cos\theta=\frac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{\sin^2\alpha+\sin^2\beta+\sin^2\gamma}, \] then what is the value of \(\theta\)?
- Consider the following two statements: Statement 1: If \(A\), \(B\), and \(C\) are points with position vectors \(\mathbf{a}=2\hat{i}+\hat{j}+\hat{k}\), \(\mathbf{b}=3\hat{i}-\hat{j}+3\hat{k}\), and \(\mathbf{c}=\hat{i}+7\hat{j}-5\hat{k}\), then the figure \(OABC\) forms a tetrahedron. Statement 2: If the position vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) of points \(A\), \(B\), and \(C\) are non-coplanar, then \(OABC\) is a tetrahedron, where \(O\) denotes the origin. Choose the correct option.
- Find the locus of a point whose sum of the squares of its perpendicular distances from the planes \(x+y+z=0\), \(x-z=0\), and \(x-2y+z=0\) equals 19.
- A moving plane always contains the fixed point \((1,2,3)\). The set of points that are the perpendicular projections of the origin onto this plane is described by
- The direction cosines \(l,m,n\) of one of the two lines satisfying the relations \(l-5m+3n=0\) and \(7l^2+5m^2-3n^2=0\) are
- A straight line is equally inclined to the x-axis and the y-axis, making an angle \(\alpha\) with each. If its angle \(\theta\) with the z-axis satisfies \(\sin^2\theta=2\sin^2\alpha\), determine \(\alpha\).
⚔️ Practice JEE Main Maths free + battle 1v1 →