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Online Practice: Scalar and vector products for JEE

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Q1MathsUnit 12: Vector Algebra
Consider ΔABC\Delta A B C with A≡(a⃗),B≡(b⃗)A \equiv(\vec{a}), B \equiv(\vec{b}) and C=(c⃗),C=(\vec{c}), ff b⃗.(a⃗+c⃗)=b⃗⋅b⃗+\vec{b} .(\vec{a}+\vec{c})=\vec{b} \cdot \vec{b}+ a⃗⋅c⃗;∣b⃗−a⃗∣=3;∣c⃗−b⃗∣=4\vec{a} \cdot \vec{c} ;|\vec{b}-\vec{a}|=3 ;|\vec{c}-\vec{b}|=4 then the angle between the medians AM‾\overline{A M} and BDB D is
Q2MathsUnit 12: Vector Algebra
The points A(−1,3,0),B(2,2,1)A(-1,3,0), B(2,2,1) and C(1,1,3)C(1,1,3) determine a plane. The distance of the plane A,B,CA, B, C from the point D(5,7,8)D(5,7,8) is
Q3MathsUnit 12: Vector Algebra
The vectors i^+2j^+3k^,2i^−j^+k^\hat{\boldsymbol{i}}+\boldsymbol{2} \hat{\boldsymbol{j}}+\boldsymbol{3} \hat{\boldsymbol{k}}, \boldsymbol{2} \hat{\boldsymbol{i}}-\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}} and 3i^+j^+4k^\mathbf{3} \hat{\mathbf{i}}+\hat{\boldsymbol{j}}+\mathbf{4} \hat{\boldsymbol{k}} are so placed that the end point of one vector is the starting point of the next vector. Then the vectors are :
Q4MathsUnit 12: Vector Algebra
In a triangle ABC,A B C, right angled at the vertex A,A, if the position vectors A,BA, B and CC are respectively 3i~+j~−k~,−i~+3 \tilde{i}+\tilde{j}-\tilde{k},-\tilde{i}+ 3j~+pk~\mathbf{3} \tilde{j}+p \tilde{k} and 5i~+qj~−4k~,5 \tilde{i}+q \tilde{j}-4 \tilde{k}, then the point (p,q)(p, q) lies on a line
Q5MathsUnit 12: Vector Algebra
Given ∣A→1∣=2,∣A→2∣=3→\left|\overrightarrow{\boldsymbol{A}}_{1}\right|=2,\left|\overrightarrow{\boldsymbol{A}}_{2}\right|=\overrightarrow{\mathbf{3}} and ∣A⃗1+A⃗2∣=3.\left|\vec{A}_{1}+\vec{A}_{2}\right|=3 . Find the value of (A→1+2A→2)⋅(3A→1−4A⃗2)\left(\overrightarrow{\boldsymbol{A}}_{1}+\mathbf{2} \overrightarrow{\boldsymbol{A}}_{2}\right) \cdot\left(\boldsymbol{3} \overrightarrow{\boldsymbol{A}}_{1}-\boldsymbol{4} \boldsymbol{\vec { A }}_{2}\right)
Q6MathsUnit 12: Vector Algebra
Assertion Let aˉ,bˉ,rˉ\bar{a}, \bar{b}, \bar{r} be the vectors such that rˉ+\bar{r}+ r‾×a‾=b‾then⁡∣r‾∣2=(a‾⋅b‾)2+∣b‾∣21+∣a‾∣2\overline{\boldsymbol{r}} \times \overline{\boldsymbol{a}}=\overline{\boldsymbol{b}} \operatorname{then}|\overline{\boldsymbol{r}}|^{2}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}})^{2}+|\overline{\boldsymbol{b}}|^{2}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} Reason r‾=(a‾⋅b‾)a‾+b‾+a‾×b‾1+∣a‾∣2\overline{\boldsymbol{r}}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}}) \overline{\boldsymbol{a}}+\overline{\boldsymbol{b}}+\overline{\boldsymbol{a}} \times \overline{\boldsymbol{b}}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}}
Q7MathsUnit 12: Vector Algebra
Equation of the plane containing the lines⁡r‾=(i‾−2j‾+k‾)+t(i‾+2j‾−k‾)\operatorname{lines} \overline{\boldsymbol{r}}=(\overline{\boldsymbol{i}}-\boldsymbol{2} \overline{\boldsymbol{j}}+\overline{\boldsymbol{k}})+\boldsymbol{t}(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}}) r=(i‾+2j‾−k‾)+s(i‾+j‾+3k‾)\boldsymbol{\boldsymbol { r }}=(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}})+\boldsymbol{s}(\overline{\boldsymbol{i}}+\overline{\boldsymbol{j}}+\mathbf{3} \overline{\boldsymbol{k}}) is

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