List of top Mathematics Questions on 3D Geometry asked in AP EAPCET

If the angle \(\theta\) between the line \[ \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} \] and the plane \[ 2x-y+\sqrt{\lambda}z+4=0 \] is such that \(\sin\theta=\frac13\), then the value of \(\lambda\) is:

If \((K,3,5)\), \((2,-1,2)\) are direction ratios of two lines and the angle between them is \(45^\circ\), then a value of \(K\) is:

If \(A(1,1,1)\), \(B(2,3,4)\) and \(C(2,5,7)\) are the vertices of \(\triangle ABC\), then the length of the altitude drawn through the vertex \(A\) is:

\(A(-4,9,k)\), \(B(-1,k,k)\), \(C(0,7,10)\) form an isosceles right-angled triangle. If \(AB=BC\) and \(AC\) is an integer then the perimeter of \(\Delta ABC\) is

If N is the foot of the perpendicular drawn from the point \(P(5,-1,3)\) to the line passing through the points \(A(1,3,-5)\) and \(B(3,-1,5)\) then the ratio in which N divides AB is

If l, m, n are the direction cosines of a normal drawn to the plane \(2x-3y+6z-7=0\) and d is the length of the perpendicular drawn from origin to this plane then \(7d|l+m+n|=\)

If $\theta$ is the acute angle between a line $\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{1}$ and a normal to the plane $2x+3y+4z=0$ then $\tan^{2}\theta+sec^{2}\theta=$

If the point P which divides the line segment joining $A(1,1,1)$ and $B(2,2,2)$ in the ratio 1: m lies on the plane $x+2y+3z-1=0$, then $m=$

The points (0, $\lambda$, 1), ($\mu$, 3, -1), ($\lambda$, 5, 0), ($\mu$, 6, $\mu$) taken in that order, form a square. If $\lambda$, $\mu$ are positive real numbers, then the length of its side is

The ratio in which the $xy$-plane divides the line segment joining the points $(2, 4, 5)$ and $(3, 5, -4)$ is:

If a line makes angles $45^\circ$ and $60^\circ$ with the positive $x$ and $y$ axes respectively, then the acute angle it makes with the $z$-axis is:

The acute angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is:

If the direction ratios of two lines are given by $ l + m + n = 0 $ and $ mn - 2ln + lm = 0 $, then the angle between the lines is:

Let \(A = (-3, -2, 7)\) and \(B = (3, 1, -2)\). Let a plane perpendicular to the line segment AB divide AB in the ratio 2:1. Then the intercept made by the plane on y-axis is

If \(d_1,d_2,d_3\) are the distances of the point \((1,2,3)\) from the \(X,Y,Z\)-coordinate axes respectively, then \(2d_2^2+d_3^2+1=\)

If \(P = (3, 12, 4)\) and \(Q\) is a point on the line \(OP\) such that \(OQ = 3\), then the sum of all the coordinates of \(Q\) is

\(A(2,1,2), B(1,0,0), C(1+\sqrt{3},\sqrt{3},-\sqrt{6})\) are vertices of a triangle. If the length of the median drawn through \(A\) is \(\lambda\sqrt{9-2\sqrt{3}+2\sqrt{6}}\), then \(\lambda=\)

If \(G(2,-1,2)\) is the centroid of tetrahedron \(OABC\) where \(O = (0,0,0)\) and \(G_1\) is the centroid of \(\Delta ABC\), then \(|OG_1| = ?\)

If the centroid of a triangle with vertices \[ (4,p,-3),\quad (-1,-1,2),\quad (3,5,-8) \] is given by the midpoint of \[ (1,4,-2) \] and \[ (q,2,-4), \] then \(p^2+q^2=\)

Suppose \((l_1,m_1,n_1)\) and \((l_2,m_2,n_2)\) are the direction cosines of two lines and \(\theta\) is the angle between them such that \[ \cos\theta=\pm(l_1l_2+m_1m_2+n_1n_2) \] Let \(A=(1,-2,3)\), \(B=(3,1,-3)\) and \(C=(-3,1,3)\) be the vertices of a triangle \(ABC\). Then \(\cos A=\)

Let \[ ax+by+cz+d=0 \] be the equation of a plane. Given that \[ 4a+4b+c=0 \] and \[ a+2b+c=0. \] Then \(d=\)

If the point \((3,4,5)\) divides the line segment joining the points \((1,2,3)\) and \((4,5,6)\) in the ratio \(\lambda:1\), then the point which divides the line segment joining the points \((3,4,5)\) and \((1,2,3)\) in the ratio \(-1:\lambda\) is:

Let the centroid of a triangle formed by the points \(A(4,x,1)\), \(B(y,-5,2)\), and \(C(7,8,3)\) be \(G(3,5,2)\), and \(CG\) meet \(AB\) in \(F\). Then \(F=\)

If \(a,b,c\) are the direction ratios of a line \(L\) and \(\ell,m,n\) are its direction cosines, then \[ \frac{a^2}{b^2+c^2}= \] is equal to:

If the vertices of a triangle \(ABC\) are \(A(1,2,3)\), \(B(h,-3,0)\) and \(C(-4,k,-1)\) and the centroid of the triangle is \(\left(5,-1,\dfrac{2}{3}\right)\), then triangle \(ABC\) is: