Projection of Vector on a line or direction MCQ Quiz in मराठी - Objective Question with Answer for Projection of Vector on a line or direction - मोफत PDF डाउनलोड करा

Calculation

Given:

Vertices: \(A(1, 2, 0)\), \(B(2, 0, 1)\), \(C(-3, 0, 2)\)

1) Find the lengths of AB and AC:

\(AB = \sqrt{(2-1)^2 + (0-2)^2 + (1-0)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\)

\(AC = \sqrt{(-3-1)^2 + (0-2)^2 + (2-0)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}\)

2) Let D be the point where the angle bisector of \(\angle BAC\) intersects BC.

By the angle bisector theorem, \(\frac{BD}{CD} = \frac{AB}{AC} = \frac{\sqrt{6}}{2\sqrt{6}} = \frac{1}{2}\)

3) Find the coordinates of D using the section formula:

\(D = \left( \frac{1(-3) + 2(2)}{1+2}, \frac{1(0) + 2(0)}{1+2}, \frac{1(2) + 2(1)}{1+2} \right) = \left( \frac{-3+4}{3}, \frac{0}{3}, \frac{2+2}{3} \right) = \left( \frac{1}{3}, 0, \frac{4}{3} \right)\)

4) Find the length of AD:

\(AD = \sqrt{\left( \frac{1}{3} - 1 \right)^2 + (0 - 2)^2 + \left( \frac{4}{3} - 0 \right)^2}\)

⇒ \(AD = \sqrt{\left( -\frac{2}{3} \right)^2 + (-2)^2 + \left( \frac{4}{3} \right)^2} = \sqrt{\frac{4}{9} + 4 + \frac{16}{9}}\)

⇒ \(AD = \sqrt{\frac{4 + 36 + 16}{9}} = \sqrt{\frac{56}{9}} = \frac{\sqrt{56}}{3} = \frac{2\sqrt{14}}{3}\)

∴ The length of the internal bisector of \(\angle BAC\) is \(\frac{2\sqrt{14}}{3}\).

Hence option 2 is correct

Explanation:

The additive inverse vector of vector pi + 2j - 3k is

-pi - 2j + 3k

So, \(\rm (\sqrt{ab}i+\sqrt b j+\sqrt pk)\) = -pi - 2j + 3k

Comparing both sides

\(\sqrt{ab}=-p,\sqrt b =-2, \sqrt p=3\)

So, p = 9, b = 4

Putting in \(\sqrt{ab}=-p\) we get

\(\sqrt{4b}=-9\)

i.e., 4b = 91

i.e., b = 81/4

Hence p = 9, b = 4, a = 81/4

Option (2) is true.

\(\therefore \) Required magnitude of projection \(\dfrac{|(2\hat{i}+3\hat{j}+\hat{k}).(\hat{i}-2\hat{j}+\hat{k})|}{|\hat{i}-2\hat{j}+\hat{k}|}\)

\(\dfrac{|2-6+1|}{\sqrt{6}}=\dfrac{3}{\sqrt{6}}=\sqrt{\dfrac{3}{2}}\)

Concept Used:

Projection of vector \(a\) on \(b\) is \(\frac{(a.b)}{|b|}\)

Calculation

Given:

\(A(0, 3, -3)\), \(B(1, 1, 1)\), \(C(2, 0, 3)\)

\(AB = B - A = (1-0, 1-3, 1-(-3)) = (1, -2, 4)\)

\(AC = C - A = (2-0, 0-3, 3-(-3)) = (2, -3, 6)\)

\(AB . AC = (1 \times 2) + (-2 \times -3) + (4 \times 6) = 2 + 6 + 24 = 32\)

\(|AC| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)

Projection of AB on AC = \(\frac{(AB . AC)}{|AC|} = \frac{32}{7}\)

\(\therefore\) Projection of AB on AC = \(\frac{32}{7}\)

Hence option 2 is correct

Answer : 3

Solution :

\(\overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

\(\overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathbf{k}}\)

Vector projection of \( \overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\) 

= \((\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}) \frac{\overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}|^2}\) 

= \((-3+6-4) \frac{(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\left(\sqrt{3^2+(-6)^2+2^2}\right)^2}\) 

= \(\frac{-1}{49}(3 \hat{i}-6 \hat{j}+2 \hat{k})\)

Concept:

Let \(\vec{u}~and~\vec{v}\) be two vectors. Then the vector component of the vector \(\vec{u}~on~\vec{v}\) is given by:

 \(\frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|}(\hat v) = \frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|^2}( \vec v)\)

Calculation:

Given:

\(\vec a = 4\hat i + 6 \hat j\) and \(\vec b = 3 \hat j + 4 \hat k\)

\(\vec a . \vec b=(4, 6, 0). (0, 3, 4)\)

\(\vec a . \vec b= 18\)

|\(\vec a\)| = √52 

\(|\vec b| = 5\) 

The component of vector a along b is \(\frac{{18}}{{25}}(3j + 4k)\)

Mistake PointsWe are asked to find the vector component of 'a' along 'b'. If the scalar component was asked, then there would have been only 5 in the denominator.

Explanation:

The projection vector of \(\rm\vec{a}\) on \(\rm\vec{b}\) is given by \(\rm\frac{\vec{a}.\vec{b}}{|\vec{b}|}\).

The correct answer is option 2.

Concept:

The projection of vector \(\rm\vec{a}\)  on \(\rm\vec{b}\) is given by \(\frac{\rm\vec{a}⋅\rm\vec{b}}{|\vec b|}\).

Calculation:

Given  \(\rm\vec{a}\) = 2î − ĵ + k̂  and  \(\rm\vec{b}\) = î + 2ĵ + 2k̂

 \(\vec a\cdot \vec b\) = ( 2î − ĵ + k̂ )⋅ ( î + 2ĵ + 2k̂ )

= 2 × 1 + (-1) × 2 + 1 × 2 

= 2 - 2 + 2

= 2

|\(\rm\vec{b}\)| = | î + 2ĵ + 2k̂ |

= \(\sqrt{1^2+2^2+2^2}\)

= \(\sqrt{1+4+4}=\sqrt 9\) 

= 3

The projection of vector \(\rm\vec{a}\)  on \(\rm\vec{b}\)  = \(\frac{\rm\vec{a}⋅\rm\vec{b}}{|\vec b|}\).

= \(\frac{2}{3}\)

The projection of vector \(\rm\vec{a}\) = 2î − ĵ + k̂ along \(\rm\vec{b}\) = î + 2ĵ + 2k̂ is 2/3.

The correct answer is option 1.

Concept -

Use of normal vector  and projection vector

projection of \(\overrightarrow{a}\) on \(\overrightarrow{b}\) is \(= | \frac{\overrightarrow{a}.\overrightarrow{b} }{|\overrightarrow{b}|}|\)

Solution - 

Normal vector to the plane to plane containing \(\hat{i} + \hat{j}+\hat{k}\) and \(\hat{i} + 2\hat{j}+3\hat{k}\) is 

\(\overrightarrow{n}=\left ( \hat{i} + \hat{j}+\hat{k} \right )\times (\hat{i} + 2\hat{j}+3\hat{k})\)

 \(\overrightarrow{n}=\left ( \hat{i} -2 \hat{j}+\hat{k} \right )\)

Projection of \(\left ( 2\hat{i} +3 \hat{j}+\hat{k} \right )\) on \(\overrightarrow{n}\) = \(= | \frac{\left ( 2\hat{i} +3 \hat{j}+\hat{k} \right ).\left ( \hat{i} -2 \hat{j}+\hat{k} \right )}{\sqrt{1+4+1}}|\)

= \(\frac{3}{\sqrt{6}}= \sqrt{\frac{3}{2}}\)

Hence the final answer is option 2.