Science (English Medium) कक्षा ११ - CBSE Question Bank Solutions

Find the equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y – z = 0.

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0

Show that the points `(hati - hatj + 3hatk)` and `3(hati + hatj + hatk)` are equidistant from the plane `vecr * (5hati + 2hatj - 7hatk) + 9` = 0 and lies on opposite side of it.

Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.

If l₁, m₁, n₁ ; l₂, m₂, n₂ ; l₃, m₃, n₃ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l₁ + l₂ + l₃, m₁ + m₂ + m₃, n₁ + n₂ + n₃ makes equal angles with them.

If the directions cosines of a line are k, k, k, then ______.

The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 is ______.

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2,  3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.

The locus represented by xy + yz = 0 is ______.

The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.

The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.

The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.

The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.

The cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.

The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hati + 2/sqrt(14)hatj + 3/sqrt(14)hatk`.

The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.

The angle between the line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.

The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.