Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 4 Addition of Vectors to help strengthen their preparations for exams.
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions
Question 1.
Show that the points whose position vectors are \(-2 \overline{\mathbf{a}}+3 \overline{\mathbf{b}}+5 \overline{\mathbf{c}}, \overline{\mathbf{a}}+2 \overline{\mathbf{b}}+3 \overline{\mathbf{c}}, 7 \overline{\mathbf{a}}-\overline{\mathbf{c}}\) are collinear when, \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non coplanar vectors.
Solution:
Let O be the origin of reference so that
Question 2.
If A B C D E F is a regular hexagon with centre o then prove that \(\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}= \overline{3 \mathrm{AD}}=6 \overline{\mathrm{AO}}\).
Solution:
Question 3.
In the two dimensional plane, prove by using vector methods, the equation of the line whose intercepts on the axes are a and b is \(\frac{x}{a}+\frac{y}{b}\) = 1
Solution:
Let A = (a,0), B = (0,b). P = (x,y)
Let O be the origin so that
Question 4.
Find a unit vector in the direction of the vector = \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathrm{k}}\)
Solution:
The unit vector in the direction of the vector
Question 5.
Find a vector in the direction of vector \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}\) that has magnitude 7 units.
Solution:
The unit vector in the direction of given
Question 6.
Find the unit vector in the direction of sum of the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\) and \( \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+3 \overline{\mathbf{k}}\)
Solution:
The sum of the vectors is
Question 7.
Write the direction ratios of the vector \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}-\mathbf{2} \overline{\mathbf{k}}\) and hence calculate its direction cosines.
Solution:
Question 8.
Consider the two points P and Q with position vectors \(\overline{\mathrm{OP}}=3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}}\) and \(\overline{\mathrm{OQ}}= \overline{\mathbf{a}}+\overline{\mathbf{b}}\). Find the position vector of a point R which divides the line joining P and Q in the ratio 2: 1 (i) internally and (ii) externally.
Solution:
i) The position vector of the point R dividing the joining of P and Q internally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}=\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})+(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2+1}=\frac{5 \bar{a}}{3}\)
ii) The position vector of the point R dividing the joining of P and Q externally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}=\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})-(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2-1}=4 \overline{\mathrm{b}}-\overline{\mathrm{a}}\)
Question 9.
Show that the points \(A(2 \bar{i}-\bar{j}+\bar{k}), \dot{B}(\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}), \mathbf{C}(3 \overline{\mathbf{i}}-4 \overline{\mathbf{j}}-4 \overline{\mathbf{k}})\) are the vertices of a right angled triangle.
Solution:
We have \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\)
∴ AB2 = BC2 + CA2 and hence a right angled triangle can be formed with the points A, B, and C.
Question 10.
Show that the points \(\mathbf{A}(2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}) \dot{B}(\bar{i}-3 \bar{j}-5 \bar{k}), C(3 \bar{i}-4 \bar{j}-4 \bar{k})\) are the vertices of a right angled triangle.
Solution:
Question 11.
In a ΔABC if \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the vertices A, B and C respectively, then prove that the position vector of the centroid G is \(\frac{1}{3}(\bar{a}+\bar{b}+\bar{c})\).
Solution:
Let G be the centroid of ΔABC and AD is the median through the vertex A.
Then AG : GD = 2: 1
Suppose \(\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OB}}=\overline{\mathrm{b}}, \overline{\mathrm{OC}}=\overline{\mathrm{c}}\) with
reference to the specific origin O.
Mid point of BC is = \(\overline{\mathrm{OD}}=\frac{1}{2}(\overline{\mathrm{b}}+\overline{\mathrm{c}})\)
Since G divides AD in the ratio 2: 1 we have
Question 12.
In a ΔABC, If ‘O’ Is the circumcentre, and H is the orthocentre then show that
Solution:
Question 13.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}, \overline{\mathbf{d}}\) be the position vectors of A, B, C and D respectively which are the vertices of a tetrahedron. Then prove that the lines joining the vertices to the centroids of the opposite faces are concurrent. (this point is called the centroid or the centre of the tetrahedron)
Solution:
Let O be the origin and G1, G2, G3, G4 be the centroids of ΔBCD, ΔCAD, ΔABD and ΔABC.
Then \(\overline{\mathrm{OG}}_1=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}+\overline{\mathrm{d}}}{3}\)
Suppose P is the point which divides AG1 in the ratio 3: 1 then
Similarly position vectors of the points dividing BC2, CG3 and DG4 in the ratio 3: 1 are each equal to \(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}+\overline{\mathrm{d}}}{4}\)
Hence the point P lies on AG1, BC2, CG3, DG4.
Question 14.
Let OABC be a parallelogram and D is the midpoint of \(\overline{\mathbf{O A}}\). Prove that the segment is \(\overline{\mathbf{C D}}\) trisects the diagonal \(\overline{\mathbf{O B}}\) and is trisected by the diagonal\(\overline{\mathbf{O B}}\)
Solution:
Question 15.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) be non-collinear vectors, if
are such that 3α = 2β then find x and y.
Solution:
Question 16.
If the points whose position vectors are \(3 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}-\overline{\mathbf{k}}, \quad 2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}-4 \overline{\mathbf{k}},-\overline{\mathbf{i}}+\overline{\mathbf{j}}+2 \mathbf{k} 4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\lambda \overline{\mathbf{k}}\) are coplanar, then show that \(\lambda=\frac{-146}{17}\)
Solution:
Let O be the origin and let A, B, C and D be the given points. Then
Question 17.
Find the equation of the line parallel to the vector \(2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and which passes through the point A whose position vector is \(2 \bar{i}-\bar{j}+2 \bar{k}\). If P is a point on this line such that AP = 15, find the position vector of P.
Solution:
The vector equation of the line passing through the point \(\mathrm{A}(\bar{a})\) whose position vector is \(\overline{\mathrm{a}}=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\) and parallel to the vector \(\overline{\mathrm{b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{b}}\) for some t ∈ R
Question 18.
Show that the line joining the pair of points \(6 \bar{a}-4 \bar{b}+4 \bar{c},-4 \bar{c}\) and the line joining the pair of points \(-\bar{a}-2 \bar{b}-3 \bar{c}, \bar{a}+2 \bar{b}-5 \bar{c}\) intersect at the point \(-4 \bar{c}\) when \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non coplanar vectors.
Solution:
The vector equation of the line joining points
and equation (5) is satisfied.
∴ The two lines intersect at the point from (1) is \(-4 \bar{c}\)
Question 19.
Find the point of intersection of the line \(\overline{\mathbf{r}}=\mathbf{2} \overline{\mathbf{a}}+\overline{\mathbf{b}}+\mathbf{t}(\overline{\mathbf{b}}-\overline{\mathbf{c}})\) and the plane \(\overline{\mathbf{r}}=\overline{\mathbf{a}}+\mathbf{x}(\overline{\mathbf{b}}+\overline{\mathbf{c}})+\mathbf{y}(\overline{\mathbf{a}}+2 \overline{\mathbf{b}}-\overline{\mathbf{c}})\) where a, b, c are non-coplanar vectors.
Solution:
Let \(\overline{\mathbf{r}}\) be the position vector of the point P the intersection of the line and the plane.
Question 20.
Prove that the vector equation of line through the points \(A(\bar{a}), B(\bar{b})\) is \(\overline{\mathbf{r}}=(\mathbf{1}-\mathbf{t}) \overline{\mathbf{a}}+\mathbf{t} \overline{\mathbf{b}}, \mathbf{t} \in \mathbf{R}\).
Solution:
Let O be the origin and P be any point on the line.
