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Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 19 Maths.

Answers (1)

Answer:

2\sqrt{2}   units

Hint:

By triangle law of addition.

Given:

\begin{aligned} &A \vec{B}=\hat{\jmath}+\hat{\imath} \\\\ &A \vec{C}=3 \hat{\imath}-\hat{\jmath}+4 \hat{k} \end{aligned}

Solution:             

                                                         

\begin{aligned} B \vec{C}=A \vec{C}-A \vec{B} &=(3 \hat{\imath}-\hat{\jmath}+4 \hat{k})-(\hat{\jmath}+\hat{\imath}) \\ &=2 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}   [ By triangle law of addition ]

B \vec{D}=\frac{1}{2} B \vec{C}    ..as D is the midpoint of

  \begin{aligned} &=\hat{\imath}-\hat{\jmath}+2 \hat{k} \\ A \vec{D} &=A \vec{B}+B \vec{D} \end{aligned}  [ By triangle law of addition ]

\begin{aligned} &=(\hat{\imath}+\hat{\jmath})+(\hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &=2 \hat{\imath}+2 \hat{k} \\ |A \vec{D}| &=\sqrt{2^{2}+2^{2}}=\sqrt8 \\ &=2 \sqrt{2} \text { units } \end{aligned}

     

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